TUTORIAL TEXT No. 1
ELEMENTARY INEQUALITIES
BY
D. S. MITRINOVIC
Belgrade, Yugoslavia
IN COOPERATION WITH
E. S. BARNES
Adelaide, Australia
D. С. В. MARSH
Golden, Colorado, U.S.A.
J. R. M. RADOK
Adelaide, Australia
1964
P. NOORDHOFF LTD - GRONINGEN
THE NETHERLANDS

© Copyright 1964 by P. Noordhoff Ltd. Groningen, The Netherlands.
This book or parts thereof may not be reproduced in any form without written
permission of the publishers.
Printed in The Netherlands.

FOREWORD
The problems contained in this series have been collected over
many years with the aim of providing students and teachers
with material, the search for which would otherwise occupy much
valuable time. Hitherto this concentrated material has only
been accessible to the very restricted public able to read Serbian*.
I greatly welcome the appearance of the Tutorial Texts based
on my problem collection. Cooperation with colleagues in
Australia and theU.S. A. and their initiative have made it possible.
For the preparation of this Text, I wish to thank E. S. Barnes,
D. С. В. Marsh and J. R. M. Radok.
D. S. MlTRINOVIC.
Belgrade, 23.III. 1964.
* Zbornik matematickih problema, Belgrade, Vol. I A962), Vol. II A960),
Vol. Ill A960), published in cooperation with D. Adamovii, V. Devide,
D. Djokovie, D. Mihailovii, Z. Pop-Stojanovic, S. Presic, J. Ul£ar.

TABLE OF CONTENTS
Introduction 7
0.1 Inequalities involving mean values 9
0.2 Bernoulli's inequality 15
0.3 Chebychev's inequality 16
0.4 Abel's inequality 18
0.5 The Cauchy-Schwarz-Buniakowski inequality ... 19
0.6 Young's inequality 20
0.7 Jensen's inequality 23
0.8 The Fejer-Jackson inequality 30
0.9 Jordan's inequality 30
0.10 Some integral inequalities 31
0.11 Some inequalities for symmetric functions 32
1 Elementary inequalities 36
2 Inequalities obtainable from functional considerations 53
3 Inequalities involving powers and factorials .... 77
4 Inequalities involving finite sums and products . . 91
5 Inequalities involving trigonometric functions . . . 101
6 Geometric inequalities Ill
7 Inequalities involving mean values and symmetric
functions 120
8 The inequalities of Cauchy-Schwarz-Buniakowski,
Holder, Minkowski and Chebychev 129
9 Inequalities involving integrals 131
10 Inequalities in the complex domain 138
11 Miscellaneous inequalities 144

INTRODUCTION
This tutorial text and problem collection is designed to intro-
introduce the student, at undergraduate or senior high school level,
to the elementary properties of inequalities. A knowledge of
algebra, geometry and trigonometry, together with a first course
in calculus, provides a sufficient background for almost all the
material. It is hoped that, as a collection of problems, the book
will be a useful adjunct to regular course work, while the provision
of an introductory survey and numerous complete solutions
will give students an opportunity for independent study.
For ease of reference, the material has been roughly classified
according to subject matter or method of proof, although a
precise classification of inequalities appears to be virtually
impossible. For a careful development of the theory of inequali-
inequalities, the reader is referred to the first two monographs listed
below and to the comprehensive bibliographies given there.
Many of the problems given in this text are to be found in the
problem pages of periodicals such as the American Mathematical
Monthly which provides an abundant source of both elementary
and advanced problems.
G. H. Hardy, J. E. Littlewood, G. Polya: Inequalities, Cambridge
University Press, London, 1934.
E. F. Beckenbach - R. Bellman: Inequalities (Ergebnisse der Mathe-
matik), Julius Springer Verlag, Berlin, 1961.
E. F. Beckenbach - R. Bellman: An Introduction to Inequalities,
Random House New Mathematical Library, 1961.
N. D. Kazarinoff: Analytic Inequalities, Holt-Rinehart and Winston,
New York, 1961.
N. D. Kazarinoff: Geometric Inequalities, Wesleyan University Press
and Random House, 1961.
D. A. Kryzhanovskii: Elements of the theory of Inequalities, Moscow-
Leningrad, 1936.
G. L. Neviazhskii: Inequalities, Moscow, 1947.
P. P. Korovkin: Inequalities, Moscow-Leningrad, 1951.

MEAN VALUES
§ 0.1 Inequalities involving Mean Values
0.1.1 For every set of positive numbers A = {av a2, . . ., «„}:
n
min {ax, a2,..., an) ^ - -
+ +
+
ax a2 an
^ (eje, . . . an)Vn
. . . +an
n
_111WJ
П
v «2, . . ., «J.
/yi
is the harmonic mean of the numbers A
1 1 1
a2 . . . anI/n their geometric mean,
their arithmetic mean and
2+ ... +an
n
2 1 1Д 2\l/2
/V+a22+ . . . +an2\m
I I their root-mean-square.
1° Consider first the inequality
An = — —— ^ (ага2. . . an)Vn = Gn. A

10 ELEMENTARY INEQUALITIES
Proof 1. For n = 2, this inequality is seen to be true and can
be written in the form (д/я1~л/я2J = 0- Assume that it is valid
for n = k, i.e.,
By induction,
^ Gh.
A — k+1 ' v ^гчс+л. > , лк-isilk _ r
Thus, it follows that
Ak+1 = \{Ak+A) ^ {AkA)V* ^ (G*GI/2 = {G^\A^\II№,
i.e.,
whence
This completes the inductive proof of the inequality A).
On the basis of this proof, it is easy to show that equality holds
in A) if and only if a1 = я2 = • • ■ = an-
This result is obvious if n = 2. Assume that it holds for some
n = k 5; 2. From the above proof, we see that if Ak+1 = Gk+1,
then
Since Ak = Gk, we have
ax = a2 = . . . = ak;
and since A = G, we have
, ai~Tcl2~T' ■ ■ ■ ~^~akl~ak+l
ak+l — Лк+1 — fc-l-l '
whence
Я1 == Я2 ^ ' • • == ^fc == ^k+1'
Finally, A) certainly holds with equality if all at are equal.
Thus our assertion is established by induction.
Proof 2. The validity of a statement P(n) can be established
by the method of regressive induction in the following manner:

MEAN VALUES 11
P(n) holds for infinitely many values of n;
P(n) implies P(n—1) for all n > 1.
Combine the methods of progressive and regressive induction x.
It has just been seen that A) holds for n = 2. Assume now
that for all positive numbers ax, a2, . . ., an, some of which may
be equal to each other, A) holds for n = 2k (where k is some
natural number). \
Form the sum
^ + «2+ ■ ■ ■ +O-V ^ = 2i:+1 = 22к = 2n^
V
or
2 \
2
From the inequality
я1 + я2 ,
^ Vaxa2 {ax, a2 > 0), B)
we obtain
an+i+an+2+ ■ ■ ■ +«2«\
_1 (a1+a2+ ■ ■ ■ +a
2 \ n
• • • +an an+i+an+2+ ■ ■ ■ +«2n\2
n n
Using the assumption that the inequality A) is valid for n = 2k,
the last inequality gives
a1+a2+ . . . +a2n aVln(a a a VlnV-l*
g ^ Ц«1«2 • • • <*n) ' {"■n+lan+2 ■ ■ ■ a2n) S
= {axa2 . . . a2nyi*\
Thus, the relation A) holds for every ne{2, 22, 23, . . .}. Assume
now that the inequality A) has been proved for n, and replace
in it an by (^+«2+ . . . +an_1)j(n—l). Then
1 Alternatively, ascending and descending induction.

12 ELEMENTARY INEQUALITIES
i
«i+a2 + • • • + an-\ + ■
n—1
C)
When the left-hand side is simplified this relation assumes the
form
whence
and, finally,
a1+a2+ . . ■
»—1 - v - -
Thus it follows from the assumption of the truth of the relation
A) for я that it also holds for я—1.
This completes the second proof of A).
2° On the basis of the inequality A) of 1°, we have for the
numbers 1/я1; 1/я2, . . ., 1/я„,
1 1 1
— • (!)
*.«! a2 aj n
This inequality becomes an equality if
a,x = a2 =... = «„. B)
It follows from A) that
foe, . . . anfi\ C)
1 1

MEAN VALUES 13
Consider now the relations:
. . . +an_1an), D)
2aiaj ^ а?+а}2, since (яг-я^J ^ 0. E)
Replacing 2aiai by a?-\-at2 on the right-hand side of the iden-
identity D), we obtain the inequality
(<*!+<*„+ . . • +«J2 ^ n{a1*+a2*+ . . . +an*) F)
which holds for all real at. If all the at are positive, it follows from
F) that
whence
It will now be shown that
min (аъаг, . . ., an) ^ . (8)
1
I
ax
1
(■
«2
1
■...-)
Without loss of generality, suppose that
0 < ax ^ a2 ^ . . . ^ an, (9)
and hence
min (au u2, . . ., aj = аг.
Using (9), the inequality (8) becomes
ал ал ал
ax a2 an
This inequality is valid, because, by (9),
«i/e»^ 1 {k= 1, 2, ...,n).
Consequently, the inequality (8) has been proved.

14 ELEMENTARY INEQUALITIES
Next, consider the inequality
(Я12 + Я22-(- . . . +Я„2\1/2
Again with the assumption (9), the inequality A0) becomes
/a2+a2+ . . . +an2\1'2
{ n ) ^
whence
This inequality is seen to hold, because, by (9),
ak^an(k = 1,2, .. .,»).
Thus, the mequality A0) is valid.
If the at are arbitrary real numbers, we have
/ 1 » \ 1/2
minflaj, |<*2|, ..., |я„|) ^ I— 2%2) ^ maxdfl^, |я2
\ » *=i /
0.1.2 Ргог)е йе inequality
«a
^max M,-A • • .,
where blt b2, . . ., bn are positive numbers.
If m is the smallest and M the largest of the numbers
ajbv a2/b2, . . ., ajbn,
then
m es aJb-L ^ M, m ^ a2jb2 ^ M, . . ., m ^ яи/6и ^ М,
whence it follows that
n i n
^h^laic^M^h; i.e., т^^
k=l fc=l Jc=l lc=l
as was to be proved.

BERNOULLI'S INEQUALITY 15
0.1.3 Mean values are important in probability theory,
mathematical statistics, and the mathematical analysis of experi-
experimental data.
§ 0.2 Bernoulli's Inequality
If h > —1 and n is a natural number, then Bernoulli's in-
inequality states that
{l+h)n ^ 1+nh. A)
Proof. Apply the method of mathematical induction. If n = 1,
the relation A) holds as an equality. Suppose that the relation A)
holds for n = k (}t 1), i.e. that
^ 1 + kh. B)
Multiplying this inequality by l-\-h (> 0), we obtain
A+*)*+! ^ (l + h){l+kh) = l+{k+l)h+kh*,
whence
C)
Since B) implies C), the proof is essentially complete.
Next the generalized Bernoulli inequalities, required in differ-
differential calculus, will be established:
(l+x)a > 1+ax (—1 < x Ф 0, a > 1 or a < 0), D)
(l+x)a < 1+ax (-1 < x ф 0, 0 < a < 1). E)
In fact, using Taylor's formula, we obtain
{l+x)a-l-ax = a(a1)x (\ + вху-г (О<0<1). F)
Since, by assumption, 1-\-вх > 0, this gives
sgn {{l+x)a-l-ax} = sgn {e(e-l)} {x
whence the inequalities D) and E) follow.

16 ELEMENTARY INEQUALITIES
§ 0.3 Chebychev's Inequality
If
Я\ ^5 «2 = • • • = Яп an<i ^1 ^ &2 = ■
then Chebychev's inequality states:
Proof 1. Let
fc=l k=\
then
Hence
/I V
By (i),
(ар—а^фр-К) ^0 (,м, v = 1, 2, . . ., я),
whence follows Chebychev's inequality,
я 2 «&-2й 26 ^ °-
If and only if
ax = a2 = . . . = an or Ъх = b2 = . . . = bn,
the equality sign holds in B).
Proof 2. Let
ax = fiv «2—«l = i>2, ■ ■ •• an—an~i = pn,
К = 9v K~\ = q2, . . ., bn—bn_1 = qn,

CHEBYCHEV'S INEQUALITY 17
with pv ^ 0, qv S: 0 (v = 2, 3, . . ., n), whence B) may be written
n / v - / v \ / n \ / n
as
or
Since
(w+1— i)(n+l—j) = [я+l— max (г, /)] [w+1— min(«, /)],
the last inequality becomes
n
2 [я+1-тах (», /)] [min(t, ?)-1]/>г-^ ^ 0. C)
i,}=2
Since
n-\-\— max (г, /) > 0, min(«, /) — 1 > 0, ft, q1 ^ 0
for i, j = 2, 3, . . ., n, the inequality C) is obvious.
This second proof is due to D. Djokovic.
Generalization. If
0 ^ cx ^ c2 ^ . . . ^ cn,
then
n n
Example. If a, b, c, are positive numbers and if и is a natural
number, then
(a+b+c)n ^ 3n~1(an+bn+cn).
Remarks. 1° From C), it follows that equality holds in B)
if and only if
ax = a2 ==... = an or 6X = b2 — . . . = bn.

18 ELEMENTARY INEQUALITIES
2° In the generalized Chebychev inequality, the conditions
0 rg ax, 0 ^ bv . . ., 0 ^ cx are essential. For instance, if ax = 1,
a2 = 3, bx = 1, b2 = 3, cx = —4, c2 = —3, we have
cx + c2 = _ _ = яЛ^
2 '
In many books (Cf., for instance, J. W. Archbold: Algebra, London
1958, pp. 51 — 52), the conclusions corresponding to 1° and 2° above are
incorrect. See also: C. V. Durell-A. Robson: Advanced Algebra, vol. Ill,
London 1948, pp. 370 — 371.
§ 0.4 Abel's Inequality
If K, a2, ...,«.} and {bx, b2). ..,bn} (bx ^ b2 ^ . . . ^ bn ^ 0)
are two sets of real numbers and if M and m, respectively, are
the maximum and minimum of the numbers
1» S2, . . ., Sn I Sk — 2, av\ >
\ v=l I
S
then
mbx ^ axbx+a2b2+ . . .+anbn <L Mbx.
Proof. The sum
(n \
may be written in the form
n
I>A = S161+(s2-s1N2+ . . . +(sn—sn_1)bn
= sx{bx—b2)+s2{b2—b3)+ ... +sn-x{bn-1—bn)+snbn.
Since M — тахE:, s2, . . ., sn), i.e.,
s1^M,s2^M,...,sn^M,
and, by hypothesis,
bx-b2 ^ 0, b2-bs ^ 0, . . ., bn_x-bn ^0,bn^ 0, A)
we have the sequence of relations
sx(bx-b2) ^ M(bx~b2), s2(b2~b3) ^ M(b2-b3), ...,
sn_x(bn^-bn) < M(bn_x-bn), snbn £ Mbn.

abel's inequality 19
Adding these expressions we obtain
s1(b1-b2)+s2(b2-b3) + . . . +Sn_1(bn_1-bn) + snbn ^ Mbx.
Consequently, one arrives at the inequality
v=X
B)
Since m = min(s1, s2, . . ., sn), it follows that m fSL sv . . .,
m ^ sn, and so from A),
m(bx—b2) ^ s1(b1~b2), . . ., m{bn_x-bn) ^ sn_1(bn^1—bn),
mbn ^snbn.
Hence, summing once again, we obtain
n
mbx^^avbv. C)
This result completes the proof of the double inequality
known as Abel's inequality.
§ 0.5 The Cauchy-Schwarz-Buniakowski Inequality
Consider the two sets of real numbers
A = {«!, a2, . . ., an), В = {bv b2, . . ., bn}
and form the polynomial in x:
(a1x+b1J+ (a2x+b2)*+ ...+ {anx + bn)\ A)
which is equal to
« + «22+ . . . +an*)x* + 2(a1b1+a2b2+ . . . +anbn)x
+ (V + &22+-..+O- B)
Since the quadratic trinomial B) is the sum of squares of real
numbers, its value is non-negative for all values of the variable x;

20
hence
i.e.,
ELEMENTARY INEQUALITIES
-{a1b1+a2b2+ . . .
\fc=l / U=l
^ 0,
ifc=l
C)
D)
which is known as the Cauchy-Schwarz-Buniakowski inequality.
Equality holds in D) if and only if ax = rbv a2 = rb2>... > an =
rbn, where r is a constant of proportionality.
§ 0.6 Young's Inequality
Let f(x) be a continuous function on the interval [0, с] (с > 0)
which is strictly increasing over this interval. Further assume that
/@) = 0, a e [0, c] and b e [0, /(c)].
С
Fig. 1.
The area of the curvilinear triangle О А Р (Fig. 1) is given by the
integral ^f(x)dx, while the area of ORB is given by J*/~1(x)da;,
where f~x{x) is the inverse of f(x). On the basis of Figs. 1 and 2,
we deduce the inequality
Jo Jo
A)

YOUNG S INEQUALITY
21
known as Young's inequality.
If and only if b — {{a), equality will hold.
О
Fig. 2
Examples. Г The function f(x) = хв~х{р > 1) for all x (> 0)
satisfies the conditions under which Young's inequality is valid.
In this case, the inequality A) becomes
whence
Га ГЪ
Jo Jo
p-\
^ ab,
— a?
p p -
This inequality is usually written in the form
11/ 1 1 \
— a"-\ bQ>ab ta.b > 0, p> 1, 1 =1 . B)
P q ~ \ ~ P q I W
2° The function log(l+x) also satisfies the conditions of
Young's inequality.
In this case, the inequality A) becomes
-ь
2S ab,
Га fb
log(l+x)da; +
Jo Jo
whence
»-6) ^ ab (a, b ^ 0).
C)

22 ELEMENTARY INEQUALITIES
3° Starting from the inequality B), we may derive Holder's
inequality
where
In B),
to obtain
Summing
A
replace a
.=„/
1 a/
p " p +
now over
1
«* j
and 6
1 b
n
v from
>
(I».
^>
by
v"
1 to
\l/e
■•) :
1, llP+llq =
avb
1 /
J, 1 1
/ \
w we obtain
■^
fc=l
К,
= l.
li6i
\l/e
■•)
P я "/v,,\v7x
whence Holder's inequality follows directly, because one may set
llP+l/q= 1.
Equality holds in D) if
&„ = a/ (v= 1,2,. ..,n).
4° Holder's inequality D) may be used to derive Minkowski's
inequality
n \i/j, in \i/v in u
IK+У ^ IV + IV
Since

jensen's inequality 23
we have (writing 2 f°r 2fc=i)
2 («*+&*)' = 2 ^(«к+й*)-1 + 2 ^(«fc+й*)-1- (в)
We now apply Holder's inequality to each of the sums on the
right-hand side obtaining
Inserting these inequalities in F), and noting that q(p—l) = p,
we obtain
2 («*+»*)• ^ [(SVI" + Bh'I"] B K+W9-
Division of this inequality by B (я*+^*)а)I/в now gives E).
Note. If ^) = 1, E) is an equality. If, however, 0 < p < 1, the
inequality sign in E) is reversed.
§ 0.7 Jensen's Inequality
Definition: A function f(x), defined on an interval [a, /3], is
said to be convex on this interval if for every a, b e [a, /3]
A)
Theorem: JPor every function f(x), convex on the interval [a, /3],
blMTl/W ()
tl v=,i / П „=i
я„ е [a, /3] aw^ w is a natural number.
Proof. Assume that the inequality B) is valid for some natural
number n = 2k, i.e., that
■ ■ ■ +an\ ^ /Ю+/Ю+ • ■ ■ +/K
\ ^ /Ю+/Ю+ • ■ ■ +/K) =
/ n
Consider now

24 ELEMENTARY INEQUALITIES
\ 2n / V
i.e.,
Using the definition A) of a convex function, and the relation
C), we obtain
n
£/(«,) -
2
n
+ 1JM
2n
2/K)
v=l
~ 2n
Since the inequality B) holds for n = 2k+x whenever it holds
for n = 2k, and since it holds for n = 2 (i.e., for k = 1), one
arrives at the conclusion that it is valid for every k e N.
Thus, the inequality B) is valid for infinitely many numbers
n e {2, 22, 23, . . .}.
Next it will be shown that the assumption that the inequality
B) is valid for some n implies its validity for n—1.
Assume B) to be valid for some natural number n and for
every av e [a, /3]. In B), replace the variable an by (^+#2+ . . . +
en_i)/(«— 1), whence B) becomes
/«x

jensen's inequality 25
The left-hand side of this inequality can now be rewritten as
n—1
whence D) assumes the form
/ax+a2+ . . . +an_x\
and it follows that
K) + . . . +f(an_x)}
+ . . . +ап_Л
n—1
n—1
Accordingly, if B) is valid for n, it is also valid for n—1.
Thus, under the above conditions, the inequality B) has been
proved by the method of regressive induction.
Geometric Interpretation.
Consider xx, x2, x3 e [a, /?], where xx < x2 < x3, and the
corresponding functional values f(x1), f(x2), f{x3). The area of the
triangle MXM2M3 with coordinatesMx{xx, f{xx)), M2(x2, f(x2)),
M3(x3, f{x3)) is given by ±P, where
xx f{xx)
P=\ x2 f(x2)
The function shown in Fig. 3 is convex, whence P > 0, while
the function in Fig. 4 is concave and P < 0.
m;
Fig. 3
м'2
Fig. 4
м;

26 ELEMENTARY INEQUALITIES
The condition P > 0 corresponds to
x3 f(x3)
> 0,
i.e.,
i.e.,
> О,
<
*i) +
since, by assumption, x3—x1 > 0.
If xx == я, x3 = 6, x2 = |(й+6) (й, 6 e [a, /3]), the preceding
inequality becomes
because
■''З Xl
3 Xl
In a similar manner, it may be shown that concave functions
(Fig. 4) satisfy the inequality
For a convex function, points on a chord lie above the corre-
corresponding points on the arc of the curve у = f(x) (see Fig. 5),
Fig. 5
Fig. 6
while for a concave function points on a chord lie below the
corresponding points on the arc of the curve (see Fig. 6).

JENSEN S INEQUALITY 27
Examples. Г Consider the function f{x) = xk, where k is a
natural number > 1, and show, first of all, that it is convex
for x ^ 0.
For k = 2,
Л x12 + x22 + 2x1x2
)= 1 {xltx2e[0, +00)).
Since 2xxx2 ^ x12-\-x22, the preceding inequality implies that
(Xl + xt\ V+s22
;l 2 )= 2 '
i.e.,
/si+s»\'
This shows that the function xk is convex for ж 2: 0 and & = 2.
Suppose now that the function хк(х ^ 0) is convex for
k = r (r any positive integer), i.e.,
Multiplication by the positive quantity \{хх-\-х2) now brings
the inequality B) into the form
x1+xa\'+1
=
4
It follows from the identity
x1rx2+x1x2r+(x1—x2)(x1r~x2r) = x{+1+x2T+1
that ,
because хг—х2 and x{—x{ have the same sign.
Introducing this result into C), we obtain
(-у | /v \ Г+1 -, Г 4-1 I M Г~|-1.
"^-j ^ 2^

28 ELEMENTARY INEQUALITIES
This completes the proof that the function xk (k a natural
number > 1, x ^ 0) is convex.
On the basis of our theorem on convex functions, we may
write the inequality
/*1+*я+...+з„\*31*+а:я*+. ••+*»*,
п
+ж„\
where equality holds for xx — x2 = . . . = xn.
2° We will show that on the interval [0, л] the function
f(x) = sin x is concave.
For
which may be written in the form
cQs
cos
sin
2 — sin
it follows that
f{xx)+f(x2) . xx+x2
< sin ,
2 ~ 2
because for the range of xx and x2 under consideration
0 ^ соэ!^—x2) ^ 1.
Using this fact and the properties of concave functions, we arrive
at the relation
/1 n \
(^ = xi> X2> ■ ■ ■> xn = n)>
/1 n
^ sin I— 2
\n „=i
where equality holds if and only if xx = x2 = . . . = xn.
Sufficient conditions for convexity. The following theo-
theorem permits one to determine readily whether a function is
convex (concave) over some interval.
Theorem. // the function f(x) has a second derivative f"(x)
over the interval [a, /3] which satisfies the inequality
f"(x)^0 (s 6 [a,/J]), A)
then the function f(x) is convex over the interval [a, /3].

Jensen's inequality 29
Proof. Let xx and x2 be any two points on the interval [a, /3].
Applying Taylor's formula in the neighbourhood of the point
\), we obtain
From B) and C), it follows that
+ T6(*2-*lJ [/"(£)+/"MJ- D)
It follows from A) that /"(I) ^ 0 and f"(rj) ^ 0, whence we
conclude from D) that
i.e., that the function f(x) is convex on the interval [a, /?], as was
to be proved.
Since f(x) is concave if and only if —f(x) is convex, we deduce
that, if f" (x) ^ 0 over the interval [a, /3], then the function f{x)
is concave over this interval.
Examples. Г Since {xk)" = k{k-l)xk-2 ^ 0 for k ^ 1 and
x ^ 0, the function xk[k ^ 1) is convex for x Sg 0.
2° Since (sin ж)" = — sin x ^ 0 for x e [0, jz], the function
sin x is concave over the interval [0, n].
3° Since (logx)" = — 1/x2 < 0 for x > 0, log x is concave on
@, oo). It follows therefore that, if ak > 0 (k = 1, . . ., n),
I i n \ in
log — £ «fc ^ — £ log afc = log (ага2 . . . an)Vn.
\Пк=1 / П jfe=1

30 ELEMENTARY INEQUALITIES
Since log x is increasing, this gives
This is an alternative proof of the inequality of the arithmetic
and geometric means (§ 0.1).
§ 0.8 The Fejer-Jackson Inequality
L. Vietoris г has proved the inequalities:
re re
2 Яд. sin &x > 0, ^akcoskx>0 @ < x < л), A)
provided that
«o ^ «i ^ • • • ^ «n > 0, B)
-2* (l^ft^i«). C)
If я0 = 1 and «j. = \\k (k = 1, 2, . . ., и), the conditions B)
and C) are fulfilled and the inequalities A) become
« :
£—-sin fee >0 @ < x < тг), D)
re ]^
1+2-T"coste> ° @ < x < я). E)
*=i «
The inequality D) is known as the Fejer-Jackson inequality.
§ 0.9 Jordan's Inequality
Since sec2 0 2; 1 for 0 sS в < \л, we have, integrating over
[0, 6],
tan в ^ в for 0 ^ в < \n.
1 tlber das Vorzeichen gewisser trigonometrischer Summen (Sitz. Ber.
Oil. Ak. Wiss., Bd. 167, 1958, S. 125—135; - Anzeiger Ost. Ah. Wiss.,
1959, S. 192—193).

INTEGRAL INEQUALITIES 31
Hence
d /sin 0\ cos в ,
deVT-j^-w- F~tan e) = °for °<e<^-
Thus, sin в/в is continuous and decreasing on @, \л\, whence
sine 2
2> — for 0 < в ^ \n.
в л
This inequality is known as Jordan's inequality. Equality holds
only when в = \n.
Combining this with the inequality
sin в ^ в (б ^ 0),
we deduce that
§ 0.10 Some Integral Inequalities
The inequalities of Cauchy-Schwarz-Buniakowski and Holder
(§§ 0.5, 0.6) have integral analogues, which may be proved by
similar arguments. All functions occurring here are assumed to be
integrable over an interval [a, /3].
Schwarz's Inequality
Г f(x)g(x)dx 2 ^ f \j{x)\*Ax\ |g(x)|2dx. A)
J a J a J a
Proof. For all real t,
and so
f
J a
t2[ f2(x)dx+2t{ f(x)g{x)dx+ [ g2{x)dx ^ 0, all t.
J a J a J a
This quadratic in t therefore has a non-negative discriminant,
from which A) follows immediately.

32 elementary inequalities
Holder's Inequality
J f(x)g(x)dx fijj
1/Q
Proof. We again use the inequality B) of § 0.6, namely,
ab ^ — a" H b" (a, b ^ 0, p > 1, — + — = A. C)
P q \ P q I
Set
1/(х)Г(
1/9
and replace a, b in C) by |/(х)|/Л, |g(x)|/.B, respectively. We
obtain
^l/(M)I^J/()r J(m
whence by integration
If
|/(a;)g
BJa
ABJa
Thus
qB"
11
— + —
p q
f f(x)g(x)dx
«'a
and this is Holder's Inequality B).
§ 0.11 Some Inequalities for symmetric functions г
1. Consider the function
where a, b, c, p (^ 1), qBz 1), r(^ 1) are real constants and
а ФЬ Ф сф a, and its first derivative
1 D. S. Mitrinovic: О nekim nejednakostima, Publikacije Elektro-
tehnickog fakulteta и Beogradu, ser. Matematika i fizika, No. 29 A959),
1-4.

SYMMETRIC FUNCTIONS 33
f'(x) = {х-а)*-1 {x-b)Q-1{x-c)r
r{a+b)]x+{pbc + qca+rab)}.
Without loss of generality, it may be assumed that я < b < c.
The function f(x) has the three zeros x = a, x = b, x = с
By Rolle's Theorem, the derivative has a zero between a and b:
Another zero occurs between b and c:
P 2(p+q+r)
+ [(p(b+c)+q{c+a)+r(a+b)Y
— ±(p+q+r){pbc+qca+rab)]V2}.
Since these "zeros of the derivative, which lie in the intervals
(a, b) and (b, c), are real and distinct, we have
[p(b + c)+q(c+a)+r{a+b)]2~±(j>+q+r)(pbc+qca+rab) > 0.
We also obtain the inequality
min(a, b, c) < a < med(a, b, c) < /3 < max (я, Ъ, с),
where med(a, b, c) denotes that one of the numbers a, b, с which
lies between max(a, b, c) and min(a, b, c).
2. Consider the function
g(x) = {x—a)(x—b){x—c){x-d)
= xi- B a)xs+ B ab)x2- QT abc)x+abcd,
where a, b, c, d are distinct real numbers, and its first derivative
g'{x) = 4x3-
where 2 я> X a^> 2 a^° are *ne elementary symmetric functions
of the variables a, b, c, d of degrees 1, 2, 3, respectively. All three
zeros of the third order polynomial g' (x) are real and distinct. The

34 ELEMENTARY INEQUALITIES
condition that all three zeros of a polynomial
are real and distinct is G2+4H3 < 0, where
Я = aoa2—a12, G = a02a3—3a0a1a2+2a1s
For the above function Я and G are found to be
G = — 1G % abc+8(£ a)(£ ab)—2(£ aK.
As a result we find the inequality
or
108B л&сJ-9B я&JB яJ-108 2 « 2 я& 5>&с
+27(J яK 2 я&с+32B ei)» < 0.
Hence we arrive at the following inequalities:
2° If 2 я = 0, then 27B abcJ + 8(£ abf < 0.
2° If 2 я& = 0, then 4g я&сJ+ QT яK 2 ябс < 0.
3° If 2 лЬс = 0, then 32 2 яб —9B яJ < 0.
3. Applying this procedure for forming inequalities to the
function
h{x) = {x—ay {x-by {x-c)r [x-d)s,
where a, b, c, d are real, distinct numbers, p'Szl, q^.l,r^l
s ^ 1, one finds an inequality which contains the preceding
inequality as a special case.
The first derivative of the function h(x) is
h'{x) = {x-ay~1{x-by-1{x~c)r~1{x-d)s~1
X {p{x—b)(x—c)(x—d)Jrq(x—c)(x—d)(x—a)
+r(x—d)(x—a)(x—b)+s(x—a)(x—b){x—c)},
or

SYMMETRIC FUNCTIONS 35
h'(x) = {x-ay-1{x—b)"-1{x—c)r-1{x-d)s~1
X {{p+q+r+s)x3-[p{b+c+d)+q{c+d+a)+r{d+a+b)
+s(a+b+c)]x2+[p(cd+db+bc)+q(da+ac+cd)
+r(ab+bd+da)+s(bc+ca+ab)]x
— (pbcd+qcda+rdab+sabc)}.
The third order polynomial P(x) = a0x3+Sa1x2+Sa2x+a3
contained within the braces has three real, distinct zeros, and
hence
С2 + 4Я3<0, A)
where H = aoa2—a12, G = a02a3—Зя0я1я2+2я13.
It follows from this that
ao =
-3eL - p{b+c+d)+q{c+d+a)+r{d+a
Зя2 = p(cd+db+bc)+q(da+ac+cd)+r{ab+bd+da)
-\-s(bc-\-ca+ab),
— я3 = pbcd-{-qcda-\-rdab-\-sabc.
For instance, if
c+d)+q{c+d+a)+r(d+a+b)+s{a + b + c) = 0,
the condition A) becomes
4яо3й23+«о4йз2 < О.
whence
4й23+я0я32 < 0, because a0 = p+q+r+s > 0,
and
r(ab+bd+da)+s(bc+ca+ab)]s
— 27(p+q+r+s) {pbcd+qcda+rdab+sabcJ < 0.
If the zeros of the polynomial P(x) are denoted by xx, x2, x3
(хг < x2 < x3) and we assume that
a < b < с < d,
we have the inequalities

36 ELEMENTARY INEQUALITIES
a < x: < b < x2 < с < x3 < d,
where xlt x2, x3 are functions of the parameters p, q, r, s, a, b, c, d
which may be determined by Cardan's formulae.
§ 1. Elementary Inequalities
1.1 Prove the inequality
(и!J <kl{2n—k)\,
where n and k (< n) are natural numbers.
Proof. If the given inequality is divided by n\k\, we obtain
. . . Bn—k).
This inequality is obvious, for the products on the left- and
right-hand sides contain the same number of factors and each
factor on the left-hand side is less than each factor on the right-
hand side.
1.2 Prove that
f{x) = 2x3+3x2—12x+7 > 0,
if x > 1. For which values of x is f(x) < 0?
Hint. Express f(x) as a polynomial in (ж—1).
1.3 Prove the inequality
{2k)! < 2№{k\J {k a natural number). A)
Method 1. For k = 1, this inequality is valid. Consider any
k for which A) holds. If the inequality
k^l) B)
is true, it follows from A) and B) that
2. C)
The validity of B) may be established without difficulty, for
it is equivalent to the inequality

ELEMENTARY INEQUALITIES 37
0 < 2/b+2
which holds for k> — 1. Thus an inductive proof is easily
constructed.
Method 2. Sinceг
Bft) I = Bft)!lBft-l)!!, 22k{k\J = {{2k)\\}2,
the given inequality may be written in the form
Bft—1I1 i.i2r-l
This last inequality is obvious, because
1.4 Prove the inequality
a+&|*> ^ 2» (\a\v+\b\v) {p ^ 1).
Proof. Without loss of generality, suppose that \a\ ^ \b\.
Then
\a+b\ ^2\b\,
whence
la + SI» ^ 2*\Ъу ^ 2г'(|я
1.5 If a > & > 0, prove the inequality
\ja—^jb < ^Ja—Ъ (n a natural number > 1). A)
Solution 1. Setting я — b = с > 0, inequality A) may be
rewritten
B)
Assume that the inequality B) does not hold for some b, с
(b, с > 0), i.e., that
V^H^^6+v^ C)
Then
whence
1 «!! = n(n — 2) (я—4) . . ., where the last factor is 1 or 2.

38 ELEMENTARY INEQUALITIES
b+c ^
and
■*(Vc)*. D)
Thus, assuming C) to be true, we are led to the false relation
D). Accordingly, the inequality A) holds subject to the stated
assumptions regarding the parameters n, a, b.
(This solution is due to D. Djokovic.)
Solution 2. Consider the function
fix) = xVn— (x~l)Vn (n> l,x^l).
Then
nf'{x) = х1/"-1-(а;-1I/и-1 < 0 [x > 1).
Therefore f{x) decreases for x > 1. Hence, since /A) = 1, we
have f(x) < 1 for x > 1, i.e.,
a;V»_i < {x-\)Vn [x > 1). E)
The inequality E) is satisfied for x = а\Ъ > 1, because
a > Ъ > 0. Therefore, setting x = я/6, we have
(тГ-
Vй
i.e.,
as was to be proved.
1.6 If a > b > 0, then
ta-b. A)
Prove this inequality and find a bound for
^/anJrkn—^/bnJrkn [n a natural number, k ^ 0, я > b > 0).
Moreover, how must the inequality A) be modified when a and
b are arbitrary real numbers?

ELEMENTARY INEQUALITIES 39
Solution, Start with the identity
{x—y){xn-1+xn~2y+ . . . +xyn~2+yn~1) = xn—yn B)
and set
x = %an+kn, у = ^/bn+kn.
Since x ^ я and у ^ b (k ^ 0) and я > & > 0, we obtain
from B)
If the left- and right-hand members of the last inequality are
divided by the positive number
we obtain
For n = 2, the inequality C) becomes
Va2+k2—Vb2+k2 ^ я—b.
The inequality holds for a ^ & ^ 0, while k may be an arbitrary
real number (positive, negative or zero).
If a, b, k are any arbitrary real numbers, then
\Va2+k2~Vb2+k2\ ^ ||e|-|i||.
(This solution is due to D. Djokovic.)
1.7 If a, b, c, d are real numbers and if
ad—bc= 1, A)
prove that E = a2+b2+cz+d2+ac+bd > 1.
Solution. The expression E may be written in the form
E = i+±[(a+cJ+{a-dJ+{b+dJ+(b+cJ].
The expression in brackets vanishes if and only if
a+c = 0, a—d = 0, b+d = 0, b+c = 0, B)
i.e., if a = b = c — d=0. But if this holds, A) is not satisfied.

40 ELEMENTARY INEQUALITIES
Therefore E > 1 for all real a, b, c, d satisfying A). (See 1.23).
1.8 Solve the pair of inequalities
2x—y 2y~x
< 0, -? < 0 (х,уф 0 .
У х
Solution. Setting xjy = z, the given inequalities become
2
2*-l < 0, 1 < 0.
z
It follows from the first of these inequalities that
*<i. (i)
and from the second that
z < 0 or z > 2. B)
It follows from A) and B) that the given inequalities are
both satisfied only if z < 0, i.e., if x and у have opposite signs.
1.9 Solve graphically the inequality
||ж| —1| <: a (a ^ 0).
1.10 Show that for all a
f(a) = cos Зя+4 cos 2я+8 cos a+5 ^ 0.
Solution. f{a) = 4 cos3 a-\-8 cos2 a+5 cos a-\-l
= B cos я+lJ (cos a+l) => (Va)/(a) ^ 0.
1.11 Prove the inequality
xm2/n+xn2/m < xm+n+2T+n (ж ^ у; х, у > 0; w, n > 0).
Hint. Consider the expression
xm+n—xmyn—xnym+ym+n, i.e., (xm—ym){xn—yn).
1.12 Prove the inequality
х2"-1^ <х2п+1(х ф\; х> 0; n a natural number),
and hence the inequality

ELEMENTARY INEQUALITIES 41
xn~x -\ <xn -\ (x Ф 1; x > 0; n a natural number).
xn~1 xn
Hint. Consider the product (xin-1 — l)(x — l). — See the
preceding problem.
1.13 Prove that if я2+62 = c2+d2 = 1, then
ac+bd\ fg 1.
Hint. Write
a = cos x, b = sin x, с = cos y, d = sin y.
1.14 Given the function
f(x) = ж+logx, A)
find the values of x and h for which the inequality
F(x, h) = f(x + h)+f(x~h)-2f(x) < 0 B)
holds.
Solution. The function f(x) is defined for x > 0, and for
ж > 0, ж+й > 0, x—h > 0,
f(x, h) = log (ж+A) + log (ж—А) — 2 logx.
Since
F (x, h) =log —,
the inequality B) holds for 0 < \h\ < x.
1.15 Prove the inequalities
fBx) > f(x) > f(-x) > f(~2x) (x>0),
fBx)-f(-2x) > 2{f(x)-f(-x)} (x > 0),
where f(x) = (ax—l)lx (a > 1).
What happens if x < 0?
1.16 Prove the inequality
Va2-b2+V2ab-b2 > a @ < b < a). A)

42 ELEMENTARY INEQUALITIES
Hint. Assume that the relation A) does not hold, i.e., that
for some a, b @ < b < a)
— b2 ^ a.
1.17 Prove the inequality
a+b-B — V2)Vab ^ Va2+b* (a,b^0). A)
Solution. Suppose that A) does not hold, but that for some
a, b (a,b^ 0)
a + b—B-V2)Vab> Va*+b* (^ 0). B)
Taking the square of B) leads to
2ab— (a+b)Vab > 0
which implies (since, from B), а Ф 0 and b Ф 0)
4я262 > (а+Ь)Ы о iab > (a+b)*,
i.e.,
0 > (a-bJ. C)
This relation does not hold for any values of a, b, Hence it
follows that A) is always true.
1.18 Prove the inequality
1 + cot a <2 cot — @ < a < л).
a
Solution, cot cot a = cosec a Sg 1.
1.19 Let
\x—a\+\y-~b\ < e (a,b real; e > 0).
Does it then follow that
\xy-ab\ < (H + |&| + e)e?
1.20 For what values of x is
\ax+b\ <c (а Ф 0, с > 0)?

elementary inequalities 43
Answer.
c—b c-\-b
> x > — (a > 0),
a a
c-\-b c—b
> x > (a < 0).
a a
1.21 For what values of x is
Ф1 + 1
x
Answer.
< 1?
1 1
< x < 0 or x > (a < —1),
1-j-a I—a
x < 0 or x > \ (a = — 1),
x < 0 or x > (—1 < a < +1),
1—a
x < 0 (a ^ 1).
1.22 Let
f{a, b, c, d) = (a-b)*+ (b-cJ+{c-dJ+ {d-a)*.
If a < b < с < d, prove that
f{a, c, b, d) > f{a, b, c, d) > f(a, b, d, c).
Hint. Begin by considering the differences
f(a, b, c, d)—f(a, b, d, c),
f(a, c, b, d)—f(a, b, c, d).
Generalise to the case
f(a, b, c, d, e) = (a-b)*+ F-cJ+ {c~df+ (d-e)*+ {e
where a < b < с < d < e.
Also treat the case
f(a1, a2,..., an) = (аг—я2)а+(я2—я3J

44 ELEMENTARY INEQUALITIES
1.23 Prove the inequality
S = а*+Ь*+с*+<1*+ас + Ь<1 ^ Vs, A)
where a, b, c, d are real and ad—bc= 1.
Solution. The inequality A) follows immediately from the
identity
с d
In A), equality holds if
V3 0 d rf /з
Я2 = l-v/з ь = — — V3, с = 0 and rf = —
2 3 3
1.24 Find a natural number N such that
Solution. We must choose N so that
v :
A Cauchy approximation of the sum of the series 2^=i l/(n+n3)
by an integral gives
ж
Since log(l+x) < x (x > 0), we have

ELEMENTARY INEQUALITIES 45
Hence A) holds if iV satisfies the inequality
2iV2 — 100 —
Thus, N = 5 suffices.
1.25 Prove or disprove the following inequalities:
1°: [a] + [b] ^ [a+b] ^ [a]
2°: [a] [b] ^ [eft] ^ [a]
3°: [Va] = [VM].
4°: [Vn]2 ^ и ^ [Vn]24-2[V«],
5°: Г-^и? < n < r-^n
(In 1° and 2°, я, 6 are real numbers, in 3°, a ^ 0, in 4°, n =
0, 1, 2, . . ., in 5°, n is integral).
1.26 If a, b, t are positive numbers, show that at-\-b/t^2Vab.
1.27 By graphical methods, estimate the solutions of the
inequality
Vx+Vx— 1 > Vx+1.
1.28 Show that the number
a -A-kb
M = —-— la, b real; k > 0)
\-\-k
lies between a and 6.
1.29 Determine the region of the xy-plane in which the point
(x, y) must lie in order that its coordinates satisfy the inequality
< 0.
1.30 Find the region of the plane in a Cartesian coordinate
system whose points (x, y) satisfy the condition
\\x+a\ — \y—a\\<a (a > 0).

46
ELEMENTARY INEQUALITIES
Hint. Possible cases are
1°: x+a> 0,y~a > 0,
2°: x+a > 0, у—a < 0,
3°: x+a < 0, y—a > 0,
4°: x+a < 0,y—a < 0,
corresponding to the set of inequalities
1°: — a < x—y+2a < a,
2°: -a < x+y < a,
3°: — a <x+y < a,
4°: — a < x—y + 2a < a.
A)
B)
Is the desired region just the interior of the square in Fig. 7?
■"X
Fig. 7
1.31 Determine graphically the solutions of [ж+1| + |у—2| ^ 1.
1.32 Show that if A) a ^ 1, B) b+c < a+1, C) Ъ<с,
then D) b < a.
Proof 1. Suppose that D) is false, so that
a < b.
E)

ELEMENTARY INEQUALITIES 47
From B) and E) it follows that
b + c < 6+1
or
с < 1. F)
From C) and E) we find that a ^ с and, by F),
я^с<1=>я<1,
which contradicts A). Thus, the inequality D) holds under the
conditions A), B), C).
Proof 2. By the condition C), one has also
26 ^ b+c.
By B) and the preceding inequality, one finds
26 < a+1.
By A) and this inequality
26 < 2a => 6 < a,
which is the desired result.
1.33 Show that if x is real, the function 4жA— x)/(l+xJ
cannot assume values greater than 1/2.
1.34 For what value or values of a is the condition
(x2+ax+l)l(x2+4:X + 8) < 8
satisfied for all real ж?
1.35 Determine k such that for all real x
(x2—kx+l)/(x2+x+l)\ < 3.
Hint. The given inequality is equivalent to
— 3 < (ж2—kx+l)l(x2+x+l) < +3.
Result, ke (—5, +1).
1.36 Solve the inequality {(Зж-1)/B-ж)} x/2 > 1.
Result, xe C/4, 2).

48 ELEMENTARY INEQUALITIES
1.37 For what values of x is it true that
1.38 For what values of x is it true that
2 < (Зж2-15ж+16)/(ж2—4ж+3) < 3?
1.39 In a Cartesian coordinate system, find the region of the
plane for which
1°: xy(x*—y2) > 0, 2°: (ж2—1)(ж2—у2) < 0.
1.40 In a Cartesian coordinate system, find the region of the
plane
{(x, у) :у <х}п {{х, у):у> -\{х-Щ
i.e., find the region whose points (x, y) satisfy the conditions
у <х, у> -|(ж—3).
Similarly, find the following regions
{{x, y):y2<x}n {(x, y): x*+y* > 1},
{(x, у):уй <х}г\ {(х, у):х< у2}.
1.41 Determine the region of the xy-plane containing those
points (x, y) whose coordinates satisfy sm(x-\-y) > 0.
1.42 Solve the inequality (sin 3x)/(sin xK < 0.
1.43 Solve the inequality (tan 3x)/(tan x) > 0.
1.44 Solve the inequality cos <£+sin^>> 1 (e.g., by setting
x = cos ф, у = sin ф).
1.45 For which values of x is the inequality
sin x > 2 cos2 x—1 vahd?

ELEMENTARY INEQUALITIES 49
Hint. Establish this first for x e [0, T], where T is the funda-
fundamental period of the function sin ж—2 cos2 x-\-l.
1.46 Solve the inequalities:
1°: x+\x\ < 1, 2°: x—\x\ > 2,
3°: \x2—x\+x > 1, 4°: sina;+|sin x\ > 1.
1.47 Solve the simultaneous inequalities:
4 > Q) 8жа_2х—Ъу— 6 > 0.
1.48 If a+Ь+с = 0 and a ^ — \, b ^ —1, с ^ —1, prove
that
*** 3.
Hint. Da+l)* ^ 2a-\-l (а Зг — J). Generalize this result.
1.49 Find the region of the ху-рЫпе for which the following
inequalities are simultaneously satisfied:
x2 < ly, y2 > 5x, y2 < 8x, x2 > 2y.
1.50 Find the region of the жу-plane for which the following
inequalities are simultaneously satisfied:
xy ^ b, xy S: а, у ^ тх, у Зг kx @ < a < b; 0 < k < w).
1.51 For which values of a does the following inequality hold:
1
— 1 < — [1— а— \/A-яJ-4я2] < +1?
Result. — 1 < a < 1/3 (а ф 0).
Remark. May these results be obtained by considering the quadratic
equation whose roots are
1
2a
1.52 If a and b (ab Ф 0) are arbitrary real numbers, prove
that at least one of the following inequalities is valid:

50
ELEMENTARY INEQUALITIES
2b
Proof. Since
a+V^+2b2
2b
< 1.
26
26
= ■?.
then at least one of these factors must be less than 1.
1.53 Prove the implication
ж2—2ж+3
— < 1 => x < 0.
ж2-4ж+3 - —
1.54 Prove the inequality
p+m хг—2тх-\-фг Ь—т
-- "- {p > m > 0).
p—m ~ хг-\-2тх-\-рг ~~ ф-\-
Proof. First of all, we have
A)
v =
= — 1
1 +
B)
From B), we see that у assumes smaller values for x > 0
than for x < 0. Thus it suffices, when bounding у from below,
to examine only the case x > 0.
For x > 0, we may write in succession
2mx
m
2+/>2
1+-
+fi
y^ -i+
2p
1 +
p—m
2mx p + 1
p+m p+t
C)

ELEMENTARY INEQUALITIES 51
For x < 0, we obtain, similarly,
1 1 2mx m
ж24-*2 > — 2фх, < , > ,
2тх т -Ь—т 2 2ф
>1-" и
хг-\-фг ф ф 2тх p—m'
2/^+^. D)
ф—т ф—т
Thus, C) and D) imply A).
1.55 If 0 < a < 2c, determine whether (x+a)/(х2+ах+с2)
lies between — Bc~[-a)~1 and Bc—a)~1.
1.56 If a = cos ос, с = sin ос, б2 = sin 2oc @ < ос < тг/4), and
f(x) = (ax2+bx+c)l(cxz+bx+a),
prove that
(sec oc—l)(cosec oc+1) j£ f(x) 5S (sec oc+l)(cosec oc—1).
1.57 Find lower and upper bounds for the function
(я2—2х cos a+l)l(xz—2x cos
1.58 Determine pairs of integers, x and y, which satisfy
simultaneously
y~\x*-2x\+% > 0, y+|aj-l| < 2.
Solution. We write the inequalities in the form
\x*-2x\, 2-y > |a;-l|. A)
Since the moduli \xz—2x\ and \x—1| are non-negative, it
follows from A) that у > —^ and t/ < 2; these inequalities yield
the integral values у = 0 or у = 1.
If we substitute t/ = 0 in A), we have the simultaneous-
inequalities
\x*-2x\ <|, |aj-l| < 2.

52 ELEMENTARY INEQUALITIES
Integral solutions of the first inequality are x = 0, 2 and of
the second, x = 0, 1,2. Consequently, the common solutions
are x = 0, 2. Thus, pairs of integral solutions corresponding
to у = 0 are:
«i = 0,1/! = 0; ж2 = 2, y2 = 0.
Substituting у = 1 in A), we obtain
\хг—2х\ <|, |ж—1| < 1.
The integral solutions of the first inequality are x = 0, 1, 2 and
of the second just x = 1. Hence the third integral solution is
«3 = 1. 2/з = !•
1.59 я2+62+с2 ^ |
1.60 a{a-b){a-c)+b{b-c){b-a)+c{c-a){c-b) ^0
(a, 6, с ^ 0) (Schur).
яб ci (a-\-c)(b4-d)
1.61 1 -< v 7n ,; (a, b, c, d>0).
a+b c+d — a+b+c+d
1
1.62 1+а+а*+ . . . +an < @ < a < 1).
1—a
1.63 Va+b ^ Va+Vb (a, 6^0).
1.64 a+b—^Vab ^ Va2+62 ^ a+b (a, b ^ 0).
1.65 a+~> Va*+b>a+^- — (—)* {a,b>0).
2a 2a 2a \2aJ v ;
1.66 {a-(n+l){a—b)}an < 6re+1 (а ф b; ab > 0).
1.67 2й > я (» ^ 1).
1.68 2n> 2n + l {n ^ 3).
1.69 2n > n2 (n ^ 5).
1.70 2" > n3 {n ;> 10).
1.71 3й > я4 (» ^ 8).
1-72 |хи—an| <; wr"-1!»—a| (|ж| ^ r; \a\ ^ r).

FUNCTIONAL CONSIDERATIONS 53
1.74
1.75 (l-a)n^l-na @ ^ a < 1).
1.76 A-й)" < @ < a < 1).
1.77
1.78
1 / 1\
a)»<- 0<a<- .
1— яя \ »/
§ 2. Inequalities Obtainable from Functional Considera-
Considerations
2.1 Prove the inequality
log(l+*)<^±|| (*>0). A)
Solution. Let
Since f'{x) = -ж2/2A+жJ < 0, /(ж) < /@) = 0 (ж > 0),
which is precisely the inequality A).
2.2 Prove the inequality
+
^ (l+xI/5 (x ^ 0).
5
Does this inequality hold for negative values of ж?
2.3 Let f{t) =t--t3-\ 24sin- (t > 0).

54 ELEMENTARY INEQUALITIES
Prove that, if x > 0, z > 0, x+z < 1,
f(x+z)<f{x)+f{z).
Solution. Let g(t) = f{t)jt. For 0 < t < 1,
0.
Note first that g(t) is decreasing in the interval @,1). It
follows from x > 0, z > 0, ж+z < 1 that g(x+z) < g(x) and
g(x+z) <g(z). Therefore
xg(x+z)+zg(x+z) <xg{x)+zg(z).
Using the definition of g(t), we obtain from the last formula
f{x+z)<f (?)+№■
(Solution due to D.C.B. Marsh).
2.4 Prove the inequality
|р ^ \a\p+\b\p (O^^^l). A)
Proof. If a and b have opposite signs the result is immediately
evident. Otherwise, let t = b/a (а Ф 0); the relation A) becomes
For ф = 0 or ф = 1, this result is trivial. Consider the function
{l+t)*-l-t* @ < ф < 1)
which vanishes for t = 0 and decreases as ^increases. This yields
the inequality A) for а Ф 0. A) also holds when a = 0.
What modification should be made if p > 1 ?
2.5 Given the function
X—Я)
where
@ < \x\ < 2я),
\lim f{x)jg{x) {x = 0),

FUNCTIONAL CONSIDERATIONS 65
f(x) = 2—2 cos ж —x sin ж and g(x) = ж2—sin2»;
prove the inequality
h(x)>0 (—2n<x<2n). A)
Solution. The value of the above limit is 1/4.
The function h(x) is even and thus may be considered only for
0 < x < 2n.
First of all, we have
g(x) = хг—sin2 x > 0 @ < x < 2jr).
Consider /(ж) written in the form
f(x) = 4 sin2 (|ж) — 4(J#) sin {\x) cos (-|ж)
= 4 sin2(lx) (l- *Ж ) {х Ф л).
[ tan(|x)J
If 0 < x < ж, \x < tan (|ж); consequently,
tan (\x)
Then, for 0 < x < л, f(x) > 0.
If л < x < 2л, tan(Jx) < 0 and, consequently,
\x
2 ■ > 0 {л < x < 2зг).
Thus, for л < x < 2тг, f(x) > 0.
When ж = л, one has
/(тг) = [2—2 cos x—x sin x]x=n = 4 > 0.
This proves the inequality A).
2.6 Prove the inequality
^2/)log^^2B/-a;J @<x,y<l). A)
1 X
Proof. Consider the function
f(x) =2/log^+(l-2/)log^-2B/-a;J @<жJ/< 1),
ОС 1X

56 ELEMENTARY INEQUALITIES
where у is a parameter. Then
Since 0 < x < 1, we find
whence it follows that min f(x) = f(y) = 0.
This proves the inequality A).
2.7 Prove the inequality
pxq—qx* > p—q (x > 1), A)
where ф and q are real numbers @ < p < q).
Solution. Consider the function
f(x) = pxq—qxJ>—p+q
and its derivative
f'{x) = pq(xq-1~x1>-1).
Taking into consideration the conditions on p and q, we see
that f'(x) is positive for x > 1, i.e., the function is increasing on
the interval A, +00)- The function f(x) therefore attains a
minimum at x = 1 which implies that /mln = 0. Thus, f(x) > 0
for x > 1, which was to be proved.
2.8 Prove the inequality
(y~x)ax log a < av—ax < (y—x)ay log a (x <y, a > 1). A)
Solution. The function f{t) = a* (a > 1) satisfies the condi-
conditions for the Mean Value Theorem on the interval [x, y]. Hence
=acloga (x < с < у). B)
y—x
Since the function f(t) [a > 1) is increasing, it follows from B)
that

FUNCTIONAL CONSIDERATIONS 57
a? —ax
ax log a < < ay log a (x < y; a > 1),
y—x
which is equivalent to A). This completes the proof.
2.9 Prove the inequality
1-х ^ e~x~xa (\x\ sufficiently small).
Solution. We set f(x) = e~x-xi and determine the equation
of the tangent to the curve у = f(x) at the point @,1).
Since f (x) = — (l~[-2x)e~x~xi, the equation of this tangent is
у = 1-х.
Since
f"(x) = D%
for
we conclude that у = /(ж) is convex upwards in the neigh-
neighbourhood of @,1).
Consequently,
1—x S: e~x~x for \x\ sufficiently small.
2.10 Prove the inequality
x log x i,.
0^—^-^1 (х>0,хф1). A)
ж2—1
Using this result, prove that the integral
cmlogx f1 ж log ж
is not greater than l/Bw).
Solution. Consider the function
/(ж) = 2 log ж - —

58 ELEMENTARY INEQUALITIES
Since
2 1 /ж—1\2
fix) = 1 = — < 0,
' V x хг \ x J -
the function f(x) is decreasing and, consequently,
2 log ж <0 {x > 1),
X
2 log ж > 0 @ < x < 1).
Hence it follows that
x log x x
ж2—1
The inequality
ж log ж
= ж2-1
is obvious, because for ж > 0,
sgn (ж log ж) = sgn (ж2—1).
It follows from A) that
Г1 xm los x f1
^-de^i ж
Jo ж2—1 Jo
(Revised from a solution by Z. Pop-Stojanovic.)
2.11 For which values of x is
log a; ^ V^?
2.12 Sketch the graph of the function
Vx—л/ж—a (a > 0)
and find the values of ж for which
л/ж— Vx—a > 2.

FUNCTIONAL CONSIDERATIONS 59
2.13 Find the values of x for which
^Jx—^x—a > b (па. natural number > 1; a, b > 0).
2.14 Factor the polynomial 2Ж3—Зж2+1 and determine the
values of x for which it is positive. Hence prove the inequality
f(x) = 2 cos x+sec2x—3 > 0 @ < x < \n).
By considering the function
Г
f{t)dt @<x<
show that
2 sin ж+tan x > Ъх @ < x < \n).
Solution. First of all, we have
Hence
2a;3—Зж2+1 > 0 (x > —§• and x ф 1), A)
2Ж3—Зж2+1 < 0 (ж < —|). B)
Since 0 < cos 0 < 1 for 0 < 0 < \n, we find from A)
2 cos3 0-3 cos2 0+1 > 0 @< 0 < |тг),
or
2cos0+sec2 0—3 > 0 @ < 0 < \n). C)
Integrating the inequality C) between the limits @, x), we
obtain
2 sin ж+tan x — Ъх > 0 @ < x < |тг),
which was to be proved.
2.15 Prove
<& @<a<6). A)
Solution. Applying the Theorem of the Mean to the function
x log x, we obtain

60 ELEMENTARY INEQUALITIES
b log b—a log a = (b—«)(l+log c) @ < a < с < b),
i.e.,
e
whence A) follows immediately.
2.16 Compare the magnitudes of the functions
(V/«)V"+I and (Vn+l)Vn (n a natural number).
Solution. Since
log
{Vn)v^ >(V^+i)^о ^ > log ^, (i)
V» Vw+1
< (V^+I)^ о ^ < ^^F1, B)
V« Vm+1
we compare the functions
log Vn log л/и +1
Without difficulty, we establish that the function f(%) = (log a;) jx
is increasing for 0 < x < e and decreasing for x > e. It follows
from this result that
log Vn log л/и+1
Vn Vn+1
n+l }j
+1
V» Vn+1
since л/7 < e < л/8. For n = 1, direct calculations show that
the inequality D) holds.
From A), B), C), D), we have the inequalities
(V/«)V"+I < (л/й+Т)^ (« = 1, 2, 3, 4, б, 6),
' V" (« ^ 7).

FUNCTIONAL CONSIDERATIONS 61
2.17 Prove that ж—tanh ж > 0 (x > 0), and hence that
d /sinh ж\
— >0 (x>0).
dx \ x }
Solution. Consider the function
f(x) = x—tanh ж. A)
Since
f'{x) = tanh2 x > 0 (x ф 0), B)
the function f{x) increases for x > 0. It follows from /@) = 0
and B) that
f(x) = x—tanh x > 0 (x > 0). C)
By C), it follows from the relation
d /sinh x\ cosh x
— I I = (ж —tanh a;)
and the inequality cosh x > 0 that
d /sinh x\
— >0 (ж>0.
ax \ x /
2.18 Prove the inequality
Solution. Consider the function
/(ж) = а;1/4-2а; (ж ^ 0)
and its derivatives:
For ж = jl_, the derivative /'(ж) vanishes, while /"(ж) < 0.
Therefore / (ж) attains its maximum value of f at x = ~^.
Consequently,
which was to be proved.
For x = 0, the inequality obviously holds.

62 ELEMENTARY INEQUALITIES
2.19 Prove the inequality
cost) > Ssint (t>0). A)
Solution. Since 2+ cos t > 0 for all t, we may write A)
in the form
Since
flt) = i-_zzzj:\ >o,
/1—cos A2
the function f(t) increases as t increases. Since /@) = 0, this
establishes the inequality A).
2.20 For х{Ф \) an arbitrary positive number, prove that
1°: xp— 1 > p(x—1) (p>lorp<0),
2°: 3i>-l<p{x-l) @<p<l).
Proof. 1°: Consider the function
f{x) = x»—l—p{x — l), for which /A) = 0,f(x) = p{xv~l — \).
If p > 1, then
I < 0 @ < x < 1),
/»= =0 (x=l),
[>0 (x>l).
Thus, the function f(x) attains a minimum at x = 1.
This is the proof of the inequality 1° for p > 1.
If p < 0, then
ж»-1 > 1 @ < ж < 1),
ж»-1 < 1 (ж > 1).
Hence
■ < 0 @ < x < 1),
f(x) 1=0 (ж=1),
• >0 {x> 1).
Consequently, 1° also holds for p < 0.

FUNCTIONAL CONSIDERATIONS 63
2°: If 0 < p < 1,
x11-1 > 1 @ < x < 1),
x^1 < 1 (x > 1).
Hence
( > 0 @ <x < 1),
/» = 0 (x = 1),
( < 0 (ж > 1).
The function /(ж) attains a maximum at a; = 1.
This proves the inequality 2°.
2.21 If a and 6 (|a| ^ 1, \b\ ^ 1) are real numbers, then
Vl—я2+л/1 — 62 ^ 2Vl —{-|(a + 6)}2. A)
Solution. Assume that A) does not hold, but that
Vl-a2+Vl—b2 > 2
Then, squaring both sides, we find
Vl—a2 Vl— 6a> 1—ab,
i.e., squaring again,
(l-a2)(l-62) > (l-abf => 0 > (й-6J.
On the basis of this contradiction, we conclude that the in-
inequality A) is true.
2.22 If f'(x) is an increasing function, then
/>+1)>/(аЧ-1)-/(а;)>/'(аО. A)
Use this result to prove the inequality
Solution. By Lagrange's theorem,
f(x+i)-f(x)=f(x+e) (o<e<i).
From the hypothesis on /'(ж) and the preceding equation, it
follows that

64 ELEMENTARY INEQUALITIES
as was to be proved.
The inequality A) may be written in the form
f(x+l)-f(x) > f(x) > f(x)~f(x-l). C)
Let f{x) = fa:3/2 and write C) with x successively replaced by
1, 2, . . ., n. Sum to obtain the inequality B). (Revised from
a solution by Z. Pop-Stojanovic.) Generalize the inequality B).
2.23 Prove the inequality
1— xn >n(l—x)x{n-v/2 @ < x < 1).
2.24 Prove the inequality
n n—1
< (n a natural number, 0 < x < 1).
x~n-xn
2.25 For what values of x (> 0), ft and q is
я* log «я ^a;
2.26 For what values of x is
|^/2^l + | ? A)
Solution. If x ^ — 1, then, by Bernoulli's inequality (§ 0.2),
+!-. B)
C)
For values of x for which
n
the inequality l+z/2— a;2/8 < A + zI/2 holds.
The polynomial \-\-xj2~жа/8 assumes negative values for x
exterior to the interval
(A)

FUNCTIONAL CONSIDERATIONS 65
Since, in addition, the condition ж 5: — 1 is satisfied, we con-
conclude that
1_|_ — — — < (l+a;)i/2 for ж ^ 2(\/3+l). D)
2 8
If x 5: —1 and if x is in the interval (A), i.e., if
-1 ^ж ^ 2(V3+1), (B)
then ж+l ^ 0 and 1+ж/2 —ж2/8 ^ 0, and after squaring the
inequality 1+ж/2-ж2/8 ^ (l+жI/2, we obtain
x3 (ж — 8) ^0
which holds for
О ^ ж ^ 8. (C)
Conditions (B) and (C) jointly require that 0 ^ ж <: 2(л/з + 1).
Collecting these results, we conclude that the inequality
holds for a; ^ 0.
Since, in addition, the inequality B) holds for ж ^ — 1, we
conclude that A) holds for ж 5: 0.
2.27 Is the inequality ж2 ^ 1 + 2 log ж true for ж > 0?
2.28 Show that, if ж is real, then
Give a geometric interpretation of this result.
2.29 For what values of ж is
(а—жN—Зй(я-жM+|й2(й-жL-|й4(й—жJ < 0?
2.30 Prove the inequality

66 ELEMENTARY INEQUALITIES
Proof. Consider the function
f{X)~\b+xj {X~0)'
and its derivative
The sign of the derivative is the same as the sign of the function
b—a a+x
+l
£*;=—- +l°g i—-.
a-\-x b-\-x
Since
D)
the function g(x) is decreasing and, consequently,
g(x)>g(+co) = 0. E)
On the basis of F) and C), we conclude that the function
f(x) is increasing, whence
The inequality A) is the special case of F) obtained when x = 1.
(Proof by D. Djokovic.)
2.31 Prove the inequalities
A—Ja;2)sin x < (x—^a;3)cos x @ < x < я;),
A — l^+Jja^sin x > (aj-l^+yi^^^cos ж @ < x < n).
What relation holds between the functions
№+1
I <-""m!)sin"(I.(-"*pB
2.32 Prove the inequality

FUNCTIONAL CONSIDERATIONS 67
sin2 x
cos x < —— @ < x < n[2).
Solution. This inequality is equivalent to
f(x)>g(x) (OKxKn/2), A)
where /(ж) = sin2#/cos x and g(x) = x2. Without difficulty, we
find that
sin a;(l + cos2 x)
1 2 sin2 x
f"{x) = cosa4 H —ig"(x) = 2.
COS X COS3 X
Since t+- ^ 2 (<> 0), /"(ж) > g"{x) = 2 @ < x < л/2).
T
Using this result we find
Jo
•^0
= /(a;)-g(a!) > 0,
which was to be proved.
(Solution by D. Djokovic.)
Remark by D. Adamovic. If we consider, instead of the given in-
inequality, the equivalent inequality x < sin x/\/ cos x @ < x < л/2), the
proof is simplified.
2.33 1°: Prove the inequality
ж log a; ^x— 1 (x > 0). A)
2°: Starting with this inequality, derive the inequality
5>, log & ^ 5»g ?„ B)
i = l < = 1
& > 0, & >0 (*" = 1, 2, . . ., w) and

n n
68 ELEMENTARY INEQUALITIES
C)
i = 1 i = 1
Solution of 2°. Since pjqt > 0, we deduce from A) that
— log — Sg — — 1.
it ii it
Since qi > 0, the preceding inequality becomes, after multiplica-
multiplication by qo
Pi log j ^ Pi-ii-
Summing both sides of this inequality over i, one finds
" Pi > " _
With C), this inequality gives
2 Pi log — ^ 0 => 2 (/», log /»,-/», log &) ^ 0
n n
^ X Pi log Pi ^ ^1 Pi log ?<,
s=l f=l
as was to be proved.
Equality holds in B) if and only if pt = qt (i = 1, 2, . . ., n).
The inequality B) occurs in Information Theory. See, for example, L. Bril-
louin: Science and Information Theory, New York 1956, pp. 13—14.
2.34 Prove the inequality
0 < tyl+x — 1-^x+lx* < -^x3 (x > 0).
Proof. Consider
f(x) = -^/Т+ж-l—lx+%x* (x > 0)
and its derivatives

FUNCTIONAL CONSIDERATIONS 69
Then
/»=/'@)+ Г f"(x)dx>f'(Q)=O,
Jo
/(я)=/@)+ Г/»dz >/@) = 0.
Jo
This proves the left-hand side of the inequality.
Next, let
g(x) = l+^x-^+^-i/l+i (x > 0),
whence
and so
g"(x)=g"@)+ Г g"'(x)dx>g"@) = 0,
Jo
= 0,
f g'(
Jo
= 0-
This proves the right-hand side of the proposed inequality.
(Proof by D. Djokovid.)
2.35 If a > 0, b > 0, a + b = 1, then
1\2 / 1\2 26
+ ) +(b+) ^
Solution. It follows from 2^/ab t== a+b = 1 that
1/И)^4. A)
Furthermore

70 ELEMENTARY INEQUALITIES
and hence
From this result and A), we obtain
■H'-M
K)
i.e.,
2L ^ 25 => L ^ 4ju.
Note. The proof may be obtained in a different manner,
using the convex function f(x) = [x-\-ljxJ for a; > 0. In that
case we can prove the more general inequality
The function
x-\ I (a > 0)
for 0 < x < 1 is convex, since
= a (W —"j {аA-а;-2J+ж-4-1+4а;-2} > О
\ я/
@ < x < 1, a> 0).
Consequently, for a, 6 > 0 and a-\-b = 1,
i.e.,

FUNCTIONAL CONSIDERATIONS 71
(This solution is due to D. Adamovic\)
2.36 Prove that
b—a b—a
< tan b—tan a < @ < a < b <\n). A)
cos2 a cos2 b
Proof. By the Mean Value Theorem
tan b—tan a = (b—a) (a < в < b).
cos2 в
Since
A) follows immediately.
2.37 Prove that
1 n—\
n " 2w2 ^
1
n
A)
(n a natural number ^ 2; x > 0).
Proof. Let us assume that for some x (> 0), the inequality
^x (x>0) B)
ft
does not hold, but rather that
1
n
Then, raising both members to the nth power, we obtain
1 \и (n\ I in\ 1 /n\ 1
—x\ =1+ — x+ { —ж2 + ...+ \—xn.
n) \1/ n \2/ и2 \n) nn
C)
Since

72 ELEMENTARY INEQUALITIES
| П \ — x2 + . . . + ( ) — xn > 0 (x > 0; n an integer ^ 2),
\ 2 / w2 \n! nn
the relation C) is false and, consequently, B) is valid.
Next, consider the function
f(x) = yi+x-l--
1 ю—1
п 2w2
and its successive derivatives,
/» - —.
Since /'"(ж) > 0, /"@) = 0, /'@) = 0 and /@) = 0 for x > 0
and n ^ 2, it follows successively that /"(ж), /''(ж) and /(ж) are
increasing functions, whence
/(x) > 0 (ж > 0, и ^ 2).
This proves the inequality
1 n— 1
1+ — ж хг < yi+x (x > 0; n integral ^ 2). D)
The inequalities B) and D) together yield A).
Question. What happens ifa; = 0orw = l?
2.38 Prove that
x log x+ey~1—xy ^0 (x > 0).
Proof. Consider the function
f{y) = x logx+ev~1—xy {x > 0)
and its derivatives

FUNCTIONAL CONSIDERATIONS 73
Since f"(y) > 0 for all y, the function f(y) attains a minimum of
zero for у = \ogx-\-\ {x > 0). Consequently
f{y) ^0 (x> 0),
as was to be proved.
2.39 Prove that
x
,и+1
1— ■: гт^ < е
1+5+|. + ... + ^_\ (ж>0). A)
Proof. Consider Taylor's formula,
«" = I ~, +Rn. where Rn = -i—- x^ @ < в < 1).
»! (w + 1)!
Multiplying by e~x one obtains
n ~v\ e0-l)x
У \ \
(n ~v\ e0-l)x
У — \ -\ ж"+1. B)
Since 0 < 0 < 1 and ж > 0, 1 > eie~1)x, and thus
xn+l e0-l)x
"+1
Now A) follows from B) and C).
2.40 Prove that a function f{x) having a second derivative
has no zero in the interval [a, b) if there exists с е (a,b) such
that for every x e (a, b)
P(c)-2f(c)f"(x)<0. A)
Solution 1. Let us assume that the converse statement is
true. If у e (a, b) is a zero of f{x) and с е (я, b) satisfies the given
condition of our problem, then we have the following:
M = №+ ^p/'W+ ^r-V(*) = 0, xe(a, b),

74 ELEMENTARY INEQUALITIES
f"(x)
f(c) + <x/'(c)+<x2 —— = 0(a = y—c), which implies:
P(c)-2f(c)f"(x) ^0.
From this contradiction it follows that f(x) has no zero in
the interval (a, b).
Solution 2. From A) it follows that f(c) Ф 0. For x Ф с
we obtain
2f(c)f'(x)-2f(c)f'(c) = 2f(c)f"(£)(x-c) > (x-c)P(c)
according as л; ^ с In both cases
f'(x)dx
+ Г
J с
= i[2f(c)+(x-c)f'(c)]\
Hence, f{x) cannot vanish.
(Problem and solution 1 by S. PreSic. Solution 2 by D. Djokovic.)
2.41 \a smx-\-b cos x\ <S
2.42 cos4 ж+sin4 x ^ \.
2.43
2.44 2д/я>3 (x > 1).
2.45 0 ^ (ж+яJ/(ж2+а;+1) ^ f (а2—я + 1).
2.46 4/и-^/и ^ (х2-аI1п-(х1~аI1п @ ^ я ^ xx ^ ж2).
2.47 -^-j ^ж"+A—ж)» ^ 1 @ ^ж ^ 1; £> 1).
2.48 «/(я1/"-!) < «(я1/1—1) {a> 0;аф1;0<х <у).
2.49 1~(sma;)i' > 1~(Sma;)i'
ft—1 ft
2.50 < (m < n; x > 0 and a; =£ 1).
7% W

FUNCTIONAL CONSIDERATIONS 75
1 + 2ж+ . . . +nxn~1 1
2.51 1>— ■ — >— (a; > 0).
~ l+22a;+ • • .+п2хп-г — n K
l + 2a;+ . . . +nxn~1 1
2.52 n > —— > - (x> 0).
~~ n+{n—l)x++xn~1 ~ n K '
2.53 (A+l)cos-^- > +^ (
K+l Д
2.54 (e + a;N-* > (e—ж)^1 @ < x < e).
2.55 ж —-|a;2 < log(l+a;) < x (x > 0).
2.56 2ж/B+а;) < log(l + a;) {x > 0).
2.57 a;2 > (l+a;)log2(l+a;) (x > -1).
2.58
2.59
2.60 log(l+a;) > (arc tan ж)/A+ж) (х > 0).
2.61 2a; arc tana; ^ log(l+a;2).
2.62 (a;+l)log(a;+l)— a; log a; > 0 (x > 0).
2.63 x3 + 3a;+2+6a;logx > 6a;2 (a; > 1).
2.64 a;a|log x\ ^ l/(ae) @ < ж < 1; a > 0).
2.65 |M^
2.67 ^ ^ ^t^ {O0and^ 1).
x — 1 x+x1'3
1 a;
2.68 x< < — log(l—ж) < @ < x < 1).
x 1-х
~~2

76 ELEMENTARY INEQUALITIES
2.69 —— < log (l + -) < - (x> 0).
1 \ X/ X
2-70
Ъ 1+6
2.71 log - < log -X. (a>b> 0).
а 1+я
./л е ,> — х (.с ,> и, л; — vt, l, *, . . .).
2.73 — > 1-е-* (ж < 1; х ф 0).
1—х
\-\-x
2.74 —— > е2х @ < х < 1).
1—а;
(х \
) < 1-х (х < 1; х Ф 0).
1— х)
2.76 ае-ьв-Ье~ав < а-Ъ {а > Ъ > 0; в > 0).
1Н 1 > ехр (х, у>0).
2.78 log A + VT+1T2) < — + log x (x > 0).
2.79 — < log я—log 6 < "~ @ <b < a).
а Ъ
2.80 е» < (l+a;)^» (x > 0).
2.81 е2ж > 1/A—ж) @ < х < 1/2).
2.82 хех/2 < ех — 1 < хех {х>0).
2.83 ех*/2 cos ж < 1 @ < х ^ 4).
2.84 cosh х cos a; < 1 @ < х ^ тг/2).
2.85 0^е-* (l ——У ^— ж2 (ж > 0).
\ п/ 2и

POWERS AND FACTORIALS 77
2.86 \ехA2 — вх+х2)—\
2.87 xi + 8x + 12ж2 log x > 8х3-{-1 (х > 1).
„, 1 x 2-х
2.88
71ж ^|log Aж)| ^ (
n 2n 1—x
2.89 log A+cosa;) < log 2—Jo;2 @ < x < л).
2.90 log (l+x2) < x arc tan x < x2 (x > 0).
2.91 ту-1 < -—- < rxr-x (x > у > 0; r > 1).
x—y
2.92 ra^ < 3—^~ < rw7- (ж > у > 0; 0 < г < 1).
ж—«/
§ 3. Inequalities Involving Powers and Factorials
3.1 Prove the inequalities:
1°: nn>Bn~l)ll, 2°: (n+l)n > Bw)!! (n > 1).
Solution 1. 1°: For n = 2, the inequality
»" > Bw-l)!! A)
is valid.
Assuming that A) holds for n = k, i.e.,
A»>BA-1)!!, B)
then the inequality obtained by multiplying B) by (k+l)k+1/kk
is also valid; thus,
(A+l)»+i> BA-1)!! (A+l)»+i/A*. C)
If
BA-1)!! (A+l)*+1/ft*> BA+1)!!, D)
then
E)

78 ELEMENTARY INEQUALITIES
The inequality D) is valid. Indeed, instead of D), we may
consider
(k + l)k+1-Bk+l)kk > 0. F)
After application of the binomial expansion, the left-hand side
of F) becomes
which is positive for all natural numbers k.
From hypothesis B), i.e., f{k) > g(k), we have shown that
f(k+l) > g(k+l). Since, in addition, A) holds for n = 2, it
follows that it holds for every n e {2, 3, 4, . . .}.
The second inequality may be proved in a similar manner.
Solution 2. 1°: From (a—bJ ^ 0, it follows that
a2 ^ bBa—b), A)
with inequality if а ф b.
If we let a = n and b = 2k—1 (k = 1, 2, . . ., n), we have
successively
w2 ^ 1 • Bw—1),
n2 ^ 3- Bw—3),
w2 > Bw—1) • 1.
Multiplication of these inequalities yields
n2n > {Bи-1)!!}2=> nn > Bw—1)!!
2°: If in .A) we set я = n+l,b = 2k (k = 1, 2, . . ., и), we obtain
^ 2- 2я,
^ 4 Bn — 2),
(и + lJ > 2w • 2.
After multiplication, we obtain

POWERS AND FACTORIALS 79
(Solution by B. Mesihovid.)
3.2 Prove the inequality
> Зи(и!J (n > 1).
Hint. Use the inequality (n+l)n > Bw)!! (n > 1) of the
preceding exercise and the inequality 22и+1 > 3".
3.3 Prove the inequality
/w+l\n
n\ < I 1 (n a natural number > 1). A)
Proof. We start with the inequality
k+1
Setting A = 1, 2, 3, . . ., и—1, we obtain the inequalities:
2
/3\2
(т)
After multiplication, we obtain
(w+1)"
22 • 3 • 4 . . . n
whence the inequality A) follows immediately.
3.4 . Prove the inequality
4й Bп)!
й+1 < (и!J (и = - 3- 4' • • •)• ( )
Solution. For n — 2, the relation A) certainly holds, because
16/3 < 4!/B!J = 6.
Suppose that A) is valid for some natural number n = k (Si 2),
that is to say

80 ELEMENTARY INEQUALITIES
or, alternatively,
Bk)ll(klJ> 4k/(k+l). B)
After multiplying both sides of B) by
Bk+l)Bk+2)l(k+l)*,
we have
BA+2) !/{(£+1)!}2 > Bk +
which may be written in the form
BA + 2)! 4&+1 (к + 2)Bк+1)
{F+1)!}2 > k+2
Since
we have
B6+2) I 4*+!
The proof of the inequality A) now follows by induction.
3.5 Prove the inequality
{2n—\)\\ 1
Bя)!! < V«'
Solution. Let
г —1)!! 1 • 3 • 5 ... Bи-1)
,., -N =
Bи)!! 2 • 4 • 6 ... Bи)
From the inequalities
2A—1 2A
2k
it follows that
/b 19 4 4

POWERS AND FACTORIALS 81
2 • 4 • 6 ... Bw) /1 • 3 • 5 . . .Bи—l)Bw+l)
< 3 • 5- 7 . . . Bи+1) ~ / 2 • 4 • 6 ... Bw)
1 1
Bn+l)N nN
Hence
N2< l/n=> N <
3.6 Prove the inequality
2! 4! ... Bи)! > {(и+1)!}га (и a natural number ^2). A)
Solution. For n = 2, the inequality A) is valid. Let us suppose
now that it holds for n = k—1, i.e.,
2! 4! . . . BA—2)! > (A!)*-1.
Then,
2! 4! ... Bk — 2)! Bk)! > Bk)l(kl)k~1
because each of the factors 2k, 2k— 1, . . ., £ + 2 is greater than
k-\-l. Consequently, the inequality A) is proved by induction.
3.7 Prove that n\ > nn/2 (n a natural number > 2).
Proof. n\ = 1.2. . . n, (и!J = I2 • 22. . . пг, which may be
written in the form
(И1J = {1 • и} B(и-1)} {3(w—2)}... {r(n-r+1)} .. .{и ■ 1}, A)
where r(l rgj r ^ n) is a natural number. For all such r,
ф-r+l) ^w, since (r—l)(r—n) ^ 0. B)
If we set r = 1, 2, . . ., n successively, we obtain from B)
1 • n = n; 2(w—1) > n; . . .; r(n—r-\-l) > n; . . .; n • 1 = n.

82 ELEMENTARY INEQUALITIES
Hence it follows from A) that
(и!J > nn (n = 3, 4, . . .),
n\ > nnl2 (n = 3, 4, . . .).
3.8 Prove the inequality
2"("-i)/2 > n\ (n a natural number > 2).
Hint. Write 2"<"-1V2 in the form
.+<n-n _ 21- 22 . . 2n~1,
and use the inequality 2n > n-\-l (n 5; 2) which may be proved
by the method of complete induction.
3.9 Show that if xx, x2, . . ., xn are natural numbers satisfying
the equation
• • • -\~xn = nP (P a natural number),
then
Х1\+хл\+. . .+xn\ ^n(p\).
3.10 If both sides of the inequality
2 sin x—sin2 x ^ 1 (o A — sin жJ ^ 0)
are multiplied by
sin2™^ @ < x < л/2, т a natural number)
and if the inequality thus obtained is integrated between the
limits @, n/2), one obtains the inequality
Л7Г/2 Л7Г/2 Л7Г/2
2 sin2m x dx— \ sin2m+1 x dx < sin2-1^ dx.
Jo Jo Jo
Show that from this result we may derive
Bm) \\
\ 4m+l J (
3.11 Starting with the inequalities
sin2ra+1x < sin2nx < sin2"-^ @ < x < n\2),

POWERS AND FACTORIALS 83
integrate between the limits @, n/2) to prove Wallis' inequality
Bn)U J 1 I B»)!!
2w+l Ц2и—l)!!j я 1 Bи-1)!!
3.12 Prove the inequality
Bn)\\ }2 4
' < л <
B»+1)МBи— l)!!j 4и+1 (Bи —
Note. This stronger Wallis' formula was given by J. Gurland
in The American Mathematical Monthly, vol. 63,1956, pp. 643-645.
3.13 Prove the inequality
1 3 5 In— 1 1
2*4 6" 2w
(n a natural number). A)
Solution. For n = 1, the relation is valid. Let us suppose that
it holds for n = k; i.e., that
1 3 5 2A —1 1
f(b) ==- - - < = p(k\
n ' 2'4" 6' " " 2k - CA + 1)V* ^ ;'
Multiply both sides of the preceding inequality by the positive
number BA+l)/BA+2). Then we have
1 3 5 2A-1
2'I' 6- ' 2A '2A+2=CA+lI/22A+2
If we can show that
then the hypothesis f{k) ^ g(A) implies that f{k+l) <^
Let us suppose that the relation B) is not vaUd, i.e., that
Then we find that
1 BA+1J 1
BА+2J

84 ELEMENTARY INEQUALITIES
20k,
which cannot be valid when к > 0. Consequently, the relation
B) is valid. By induction, this proves A).
к
3.14 If я = ^ и, (nv natural numbers), then
Proof. Since every term in the expansion of («i+«a+ . ■ ■+nk)n
is positive and less than the sum of all the terms,
-П(»»Л) ^nn<> A).
ПК!)"
Equality holds in A) if n = nx.
This solution is by Chih-yi Wang (Mathematics Magazine,
vol. 31, No. 2, 1957, p. 113).
3.15 If xr = x(x— 1)(ж—2) ... (ж—r+1) (r a natural number),
ж0 = 1, prove the inequality
Bи-1)п_1Bи-3)п_зBи-5)„_5 . . . > ип_гип_3ии-5
where the products on both sides extend over all non-negative
indices of the same parity as и —1, and n > 1.
Solution. Since xT = \Jrl, if a; is a positive integer, the
products on the left and right-hand sides may be written in the
forms
/2n—l\
= { n-l)
2и —
= П

POWERS AND FACTORIALS 85
v=0
From these expressions we have
= п v ;
и—2v— I)
_ fitii1» n\n\ / 2n
~~ /=o 72Й]~! \2v+:
From the characteristic symmetry of the binomial coefficients
and the fact that the middle coefficient of an expansion is the
greatest, we have
Because B^) = B«)!/(и!J, we thus have i?/Z. < 1, i.e., L > R,
which proves the given inequality.
(This solution is by S. Presic).
3.16 Use mathematical induction to prove the inequality
n logn—n < log n\ < («+-|)log n—w+1 A)
(n a natural number > 1).
Solution. We set
u(n) = n log w—n, v(n) = (и+2"I°§ w—и+1.
Let us first prove the inequality
u(n) < log и! B)
This is valid for и = 1. Let us assume that B) holds for some
natural number n = k, i.e., that
u(k)<\ogkl; C)
then

86 ELEMENTARY INEQUALITIES
logA!+log(A+l) =
If we prove that
(k+l)]og(k+l)-[k+l) <k\ogk-k+log{k+l), D)
then we shall have established that
and thus that the relation B) is valid for n = k-\-l, if it is valid
for n = k.
Let us suppose that D) is false for some k, so that
(k+1) log(A+l)-(A+l) ^ к log k-k+log(k+l). E)
Then
k+1 I 1\*
M^^ll^l+j ^1
whence it follows that
(i)*a, F)
Thus, starting with E), we arrive at F), which is false for all k.
This proves the inequality D), and thereby B).
Next, consider the inequality
log n\ < {n+\) logn—n + 1 = v(n). G)
Forn = 2, we have log 2! < flog 2—1, since log 2 = 0.6931
Let us now suppose that G) is valid for n = к, i.e., that
logk\<v{k) (k>l). (8)
It then follows that
If we show that
(k+^)logk-k+l+\og(k+l)< (k+^)log(k+l)-(k+l) + l, (9)
then we shall have shown that the inequality G) holds for
n = k+1, if it holds for n = k.

POWERS AND FACTORIALS 87
The inequality (9) may be written in the form
or
log^^ + ^^r<°- A0)
Setting
we find that
f'(x) =
x{x+l){2x + lJ
For x > 0, it follows that f'(x) is positive, whence f(x) is an
increasing function for positive x. As x -*■ -f oo, /(ж) -»■ 0. Hence
/(ж) < 0 for x > 0.
This completes the proof of A0), and the proof of A) follows
by induction.
3.17 Prove the inequality
nn+i > (и-|_1)п (и a natural number > 2). A)
Proof. For n = 3, A) is true. Let us suppose that A) holds
for n = k, i.e., that
> {k+l)k {k > 2). B)
Since
it follows that

88 ELEMENTARY INEQUALITIES
From B) and C), after multiplying, we obtain
whence the truth of A) follows by induction.
3.18 Prove J. van Hengel's inequality
)" < nn+r (n, r natural numbers, n ^ 3 or r ^ 3).
Remark. Concerning this inequality see: Dickson: History of the
Theory of Numbers, vol. 2, p. 687.
3.19 Prove Goormaghtigh's inequality
(n-\-r)n < пг~р{п-\-р)п (n, p, r natural numbers, n ^ 3, p < r).
Remark. See: Mathesis, vol. 68, 1959, p. 374 — 375.
3.20 log«!> {n-\-\) log n—n-\ .
12и
3.21 ^(м+1) < A12233 . . . w"J/(ra2+7l> < ^-Bw-fl) (n > 1).
3.22 2n > 1 + ил/2^ (n > 1).
3.23 IJnTl < n~ljn (и ^ 2).
3.24 n\
(
3.25 (и!K ^ »» (-1-) •
3.26 и < {(»+1I+1/"(и-1I-1/»}1/2 < и+1/и (и > 1).
3.27
3.28 3711...D,-1)
5- 9- 13 ... D+1)
1 1-3...Bи—1) 1
3.29 =< v ^< (и > 1).
2У 2-4...2и л/2+1

POWERS AND FACTORIALS 89
2n
3^—1 / 1\ / 2\ / n — l\
з.зо < i+— i+— ... н
(и > 2).
4 • 7 • 10 ... 3.. , _, .,,-,-, - ,
3.31 —- > 1+4 l+i+i+ . . . -\ .
3- 5- 8 ... Cw+2) 3 \ '2131 n+1
3.32 й + 6 + с ^ ^b-c+^c-a+c^a-b (a, b, c,d> 0).
3.33 ас6й(с+^)с+й ^ ^"(я+й)"-^ (a, 6, c, d > 0).
3.34 aabb{a+b)a+b ^ (a2+^2)a+6 («, & > 0).
(.+ - @
3.35 a2a < (a+b)a+b(a—b)a-b < (a+ -) @ < b < я).
3.36 я6 + 6а > i (a, 6 > 0).
3.37 aabb > abba (a > b > 0).
3.38 рр/и>+*+'уПр+«+г)гг/{р+я+г) ^ i(^+?+r) (^( q> r natural
numbers).
3.39 A+яI-аA-аI+а< К A+яI+аA —яI-" @<я<1)
3.40 (я™-1^-1)™- • • (a-n)m\ ^
3.41 7й^1
2ra nn
3.42 — < — <3n-1 (и ^ 3).
3.43 и" ^ Bи-1)П ^ Bи-1)"/2.
3.44
n+1 )
3.45 (as'-5!')(aa+6e) > {a*— №){av+№) (a > b > 0, p > q)
3.46 (pa + qb)p+tl > (p + q)v+Q ■ avb" (a, b, p, q > 0, p ф q).
3.47 яс61-с+A-я)сA-6I-с ^ 1 @ < я, 6, c<l).

90 ELEMENTARY INEQUALITIES
I!J ^x (>0)
/X+l\x
.48 I—!—J
/a+b\ a+b
3.49 aabb > I-1-) {a, b > О, а ф Ъ).
\ 2 /
/a4 I J4 I С4ч <а+Ь+с)/2
3.50 , ^—1 ^ааЬъс° (a.b.oO).
\ a-\-b-\-c /
3.51 Fci+c^a+ia6+a6c)a+!)+c+d
(a, b,c,d> 0)
) A)
/ l\m / l\n /w!\
.52 A+-) > 1+-1 >2»(—)
\ m/ \ n/ \nnj
3.53 ( ) ()
/ l\2n / lX"-1 / l\n+1
3-54 Ы () ()
/ l\2
-54 Ы
3.55 2a6(c+l) ^ a';+1+6c+1+c(a1+1/<'+6i+i/<') (Я) 6, c > 0).
/w 1 l\3w
3.56 nn(-~) > (w!L (»>1).
\ 2 /
3.57 wl+и! > (w—1)! + (и+1)! (w-и > 1).
3.58
k=0 \«/ \П—1
3.59 Л
3.
n
3.61 I —

SUMS AND PRODUCTS 91
/ l\m / l\n
3.62 2< 1+- < 1+- <3 A< *»<»).
\ m) \ n)
(a\m
1+—
mj
-| {a> 0;m
nj
3.64
3.65
§ 4. Inequalities Involving Finite Sums and Products
4.1 Prove the inequality
11 I
l-\ —-\ —+...H -=> 2(Vn+l — l) (n ^ 1). A)
V2 л/3 Vn
Solution 1. For n = 1, the inequality A) certainly holds.
Let us suppose that A) is valid for n = k, i.e., that
-L+... + JL
Л/2 Vk
Then
1 1
Л/2 Vk
If we can show that
.2(Vk+l-l)+(llV
then the proof of A) will
Now B) is equivalent
.j > 2(л/&+1 — 1)
Vk+1
k+l) > 2(V*+2-l) (
follow by induction.
to
1
Vk+i
[k ^ 1), B)
(/) 2Vk+2.
If, on the contrary,
2Vk+l+(l/Vk+l) ^2Vk+2, C)

92 ELEMENTARY INEQUALITIES
then
^ 2y/(k+l)(k + 2) => 9 ^ 8.
Since hypothesis C) leads to a contradiction, it follows that
the inequality B) holds for k 22 1.
Solution 2. Since the function f{x) = 1/Vx is monotone
decreasing,
+ К№ or
which was to be proved.
4.2 Prove the inequality
4.3 Prove the inequality
M ' n+l и+2 ^ 2и 2 v У
Solution. For 1 ;S v ^ и, w+v ^ 2». Thus, l/(«+v) ^ 1/2и
for each of the n terms of f(n), whence f(n) > n/Bn) = 1/2.
4.4 Prove the inequality
f(n) = 1 H • • ■ H > 1 (n = 1, 2, 3, . . .).
/v ' n+l n + 2^ 3w+l V '
Solution. Let us suppose that the given inequality is valid
for some natural number n = k, that is to say
f{k) > 1. A)
If both sides of this inequality are increased by
1111
j | i_
36+2 "^ ' "

SUMS AND PRODUCTS 93
we obtain
If
2 11
r3k+l>1'
it will follow from the induction hypothesis that /F+1) > 1.
Now C) is equivalent to
112
36 + 4 >
36+2 36 + 4 > 36+3'
i.e.,
2C6+3) 2
C6+3J—1 36+3'
i.e.,
C6+3J—1 C6+3J'
which holds for all 6 2^ 1.
Since, in addition,/A) = 13/12 > 1, we conclude that f(n) > 1
for all natural numbers n.
4.5 Starting with the inequalities
A ()
prove that
m 2™ 1 m Зш+1
for m a natural number.

94 ELEMENTARY INEQUALITIES
4.6 Show that if ax ^ a2 ^ a3 ^ . . . ^ an ^ 0 and
£e»^£*, {k = 1, 2, ...,»), A)
then
2 а„2 ^ 2 6^2- B)
Proof. After multiplication by ak—ak+1, A) becomes
* к
(ak-ak+1) 2 «, ^ («*-«*n) 2 6^ (A = 1, 2, . . ., я), C)
v=l v=l
where an+1 = 0. Summing both sides of the relation C) from
k = 1 to k = n, we obtain
£ «,* ^ £ «A- D)
With Schwarz's inequality
the relation D) yields
whence follows B).
Equality holds in B) if and only if av = bv (v = 1, 2, . . ., n).
4.7 If xt ^ хк (г < k), yt ^ ук (г < k) or if xt >, xk {i < k),
У г ^ ук [i < k), then prove that Dn ^ ^n+i, where
r I r \ / r \
>r = r 2 я„*/„- B^I2 2/" •

SUMS AND PRODUCTS 95
4.8 Prove the inequality
(я ^0, n a natural number). A)
Solution. For и = 1, the relation A) becomes
я+я2 ^ я3+1(я ^ 0) оя(я+1) ^ (я2—я+1)(я+1), B)
which is vaUd, because
a sS (a — 1J+я for all я.
Equality holds for я = 1.
Let us now suppose that A) is valid for n = k, i.e., that
f{k) =a+a*+ . .. +a№ ^/^a^ + l) = g(k) (я ^ 0), C)
which may be more briefly written as
f(k) £ g(k) (a ^ 0). D)
Then
Consequently, if
к(а2к+1+1)+а№+1+а21с+2 ^ (k+l)(a№+s+l) (я ^ 0), E)
then
whenever D) holds.
The inequality E) is equivalent to
{k+l)a№+1(l—я2) ^ 1-(а2)к+1 (я ^ 0). F)
We will prove that F) is valid for a ^ 0. If 0 ^ a < 1, after
division by I—a2, F) is equivalent to
. . . +я2*. G)
For 0 sS a < 1, we have the inequalities
a2k+l < a2; . _ .;
from which follows the inequality G). For я = 1, equality holds
in G).

96 ELEMENTARY INEQUALITIES
If a > 1, then F), after division by I—a2, becomes
(k+l)a2k+1 ^ l+a2+ . . . +a2k. (8)
Since a > 1, we have the inequalities
a2k+1 > 1, a2k+1 > a2, . . ., a2^1 > a№,
whence follows (8).
Thus E) holds for all a 2g 0. This establishes the inequality A).
4.9 Prove the inequality
(k-nxJxk(l— x)n~k ^— A)
4
(и a natural number, x a real number).
Proof. We start with the binomial expansion
After differentiating with respect to t and multiplying by t,
we obtain
*=0
After another differentiation with respect to t and multiplica-
multiplication by t, this yields
(l)
Нхф\, the relations B), C), D) with t = хЦ1—х) become,
respectively,
Zk(nAxk(l-x)n-k=nx, F)

SUMS AND PRODUCTS 97
" ln\
£ &2 \xk(l — x)n-k = nx(l— x+nx). G)
=o \b I
k=o
Direct verification shows that E), F), and G) also hold for a; = 1.
If we multiply both sides of E) by пгхг, F) by (— 2nx) and G)
by 1, and add, we obtain
" ln\
2 {n2x2—2nxk+k2)xk[ (l—x)n-k = nx(l— x+nx)— n*x2,
k=0 \К '
i.e.,
£ (nx~kJxk (П\ {\—x)n-k = nx{l—x). (8)
k=0 \« /
Next, since
яA-я) = J-(*-£J,
we have
х[1—х)^Щ. (9)
From (8) and (9), the inequality A) follows.
4.10 Prove that
 Г n  n Ги"|
У —= — 5 "Г = и2~и (и а natural number),
*=i Lv^J *=i L«J
where [я] represents the greatest integer not exceeding a.
4.11 From the graph of the function xv(p > 0) deduce the
inequality
1 Гъ
a? < xv dx < bp @ < a < b)
b—aja
and hence prove the inequality
nv+1
<
P+l
4.12 Show that if a{ ^ — 1 {i = 1, 2, . . ., w),
6, ^1 (k= 1,2,...,»),
and

98 ELEMENTARY INEQUALITIES
then
W=l I \k=l
4.13 Prove the inequality
1 I- ... -I > 1 (и a natural number > 1)
n n+1 n2
Proof. Since, for n > 1,
1111 11
n+1 w2' и+2 n2>'"'n2—1 и2'
we find
11 11 n2—n
^ + ^+~I+---+~°>- + -
4.14 Prove the inequality
Л/2
Proof. If /fe < n (^ 2), then 1/V^ > Vv^- Setting ^ = 1
2, . . ., n, in succession, and adding the inequalities so obtained
we find
11 11
\/l Л/2
which is the desired inequality.
(Solution by B. Mesihovic.)
/ » \l/2 / » \l/3
4.15 " "
4.16 ПA+«1)>1+2я» K>0 or -l<a,<0;w> 1)
*=i *=i
4.17 ПA-в,)>1-2в, @< ак< 1; w> 1).

SUMS AND PRODUCTS 99
4.18 JJ (!+«*)< l/(l- 2 at) (l>fc<l,«*>o).
k=l I \ fc=l I \*=1 /
4.19 n(l-el)<l(
. *=1 / \ k=l
4.20 П (l+ak) ^ l+sn+ % + •. • + ^ (
k=i *'■ ill \ k=i
4.21
4.22 П A+«*) < (l-«)/(l-2a+an) @ < a < J).
4.23
/ " 1\
4.24 exp Ji
\fc=i«/
4'25
£»)>-i
4.27
4.28 1и3 < 2 Ъ
1 n 1 1
4 29 1 ^ v ^ i
W + l i=2 /%2 «
4.30
и+1 и+^)+1 и (n+kJ n n+p
4-31 |

100 ELEMENTARY INEQUALITIES
4-32 li<2-
4.33 Ь<У <- Х- (и -> n
и 2»-i l
4-34 Т< I T< n
4.35 «{(w+lI/"—1} < 2т<
4.36 2-
4-37
,?tom-~k 2m—
2иА+2
4.38 2и+1< 2 * о ,T <
1
4.39 - У v" >
(
^ \ 2
n 1
4.40 2 — < l°g и-
&=2
4.41
442 у
..,12 3 n
4.43 _ j 1 1_ . . . i _
n n—\ n-2^ ^ 1
И+1/1 1
2n \n n
h+--+j) (^

TRIGONOMETRIC FUNCTIONS
101
§ 5. Inequalities Involving Trigonometric Functions
5.1 Prove the inequality
sin ( £ xA
@<xk<n;k=l,2,...;v>l). A)
Solution. First of all, we have
sin^+a^) = sin хг cos #2+cos Xl sin x2
=s- |sin(a;14-^2)l < Isin»!! + |sinx2|.
Since, by hypothesis,
0 < x1 < л, О <. x2 <. л,
it follows from B) that
С sin Xi+sin x2.
B)
Accordingly, the inequality A) holds for v = 2. Let us suppose
that it is valid for some и E; 2) and consider the equality
sin (xn+1+ 2 **) = sin xn+1 cos I 2 ж*1 + cos ж»+1 sin ( 2 ж*) •
Hence
/ " \
sin \xn+1+ 2 ^fc
< I sin xn+1\ +
sina;n+1|+
by the induction hypothesis.
Since 0 < #n+1 < ж, the preceding inequality gives
/n+l \
sin 2 xk
\&=i
n+l
sin
@ < xk < n; k = 1, 2, . . .; n > 1).
This completes the inductive proof.
5.2 Compute the sums
X = 2 cos Aa;, 52 = 2 sin
(w a natural number >

102 ELEMENTARY INEQUALITIES
Prove that
Щ @ < x < Щ. A)
Answer.
Sx = {cos %(m-\-n-\-\)x sin \{m—w)#}/(sin |#),
52 = {sin \{m-\-n-\-Y)x sin \{m—w)x}/(sin \x).
Note. How does the inequality A) change if же (— оо, + °°)?
5.3 Solve the inequality sin x > sin 3x graphically.
5.4 If 0 < ak < n (k = 1, 2, . . ., n), is it true that
~£lsinah ^ sin— 2 <h\ ?
5.5 Prove the inequality
n
2 cos(a4—a,) ^ — \n (av a2, . . ., an real). A)
Proof.
/ « \2 / " \ 2
2 cos aj +2 sin at
П
(cos ai cos ^.+sin af sin at)
г<з
whence the inequality A) follows directly.
5.6 For what values of x is
sin x > 2 cos2#— 1 ?
Solution. The given inequality may be written in the form
2 sin2 x+sin x—1 > 0
or
2(sina;+l)(sinx—\) > 0.

TRIGONOMETRIC FUNCTIONS
103
Since sin x > —1 for all x Ф %nj2-\-2kn, the above inequality
is satisfied provided that
sin ж — \ > Ooxg {л/6+2кл, 5n/Q+2kn) (k = 0,±l,±2, . . .).
5.7 Show that if n is a natural number, then
sin nx
sin x
^n (x ф kn, k = 0, ± 1, ±2, . . .).
A)
Solution 1. For n = 1, the relation is valid. Let us suppose
that A) holds for n = k, i.e., that
sin kx
sin x
Consider the quotient
sm.(k-\-\)x sin kx
sin x sin x
It follows from this result that
B)
cos #+cos kx.
sin (
sin x
sin kx
sin x
|cos
kx\.
From B) and the preceding relation, it follows that
sin(.. , _,„
^ k+l.
sin x
This proves A) by induction.
Solution 2. The inequality A) may also be proved by starting
with the identities
sin 2px
sin x
= 2{cosa;+cos 3x+ . . .
and
'= l+2{cos
sin x
From these it follows that
4x+ . . . +cos 2px} {p ^ 1).

104
ELEMENTARY INEQUALITIES
sin 2px
sin x
sin x
2p+l.
5.8 Show that, if k is a natural number and if A and В are
real numbers, then
cos kB—cos kA
<L k2 (cos А ф cos B).
cos В—cos A
Proof. Starting with the identity
cos p—cos q = —2 sin ^(p-\-q)sin \{p—q),
we find that
cos kB—cos kA sin%k(B-{-A) sin%k(B—A)
cos B—cos A
cos kB—cos kA
A)
cos В—cos A
sin-K-B+Л)
sva.\k[B—A)
sin
-Л)
Since
|sin иж/sin ж| ^ и (и a natural number),
B)
the preceding inequality gives precisely the inequality A).
The inequality B) was proved in 5.7.
5.9 If A and В (cos В Ф cos Л) are real numbers and if k(> 1)
is a natural number, then
cos kB cos A— cos &Л cos .B
< k2-!.
cos 5—cos Л
Solution. From the inequality
|sin rx\ 5S r |sin ж| (r a natural number)
it follows that
A)
sin rx sm sy
rs
sin га sin sy+sin sx sin
sin x sin
<
sin x sin у
(r, s natural numbers). If we now set r = k-\-l, s = k—1,
x = $(A-\-B), у = %(A—B), we obtain the inequality A).

TRIGONOMETRIC FUNCTIONS 105
This solution was given by L. E. Bush [The American Mathe-
Mathematical Monthly, vol. 64, 1957, p. 651).
5.10 Is there an a such that
t2 ft t2 ft
1"9+5i>COS<> 1~? + - (°<*<*/2)?
Zi Zi^± Zi a
5.11 What conditions must be satisfied by the positive
numbers a and b so that the following inequality is valid:
sin x
(^О)? A)
x^(ж^О)?
a-\-b cos x
Solution. If we set bja = k (> 0), the inequality A) becomes
x
{>0). B)
x{x0).
~ 1 + Acosa; K — '
We cannot have k > 1, because when x tends to a root of
the equality
l-\-k cos x = 0,
(and such roots will exist), then the value of the function
sin a;
l-\-k cos ж
tends to infinity and B) cannot hold.
For k = 1, the inequality B) becomes
^SnX = 2 tan ix (x ^ 0),
x ^
1+ cos x
which is false. Consequently, we must take k < 1. Then B)
may be written in the form:
ж—sin x-\-k(x cos x—sin x) 5: 0,
x cos x—sin x 1
(x)= : ^-- (x^0).
x—sm# A
Here we take /@) = lim f{x).

106 ELEMENTARY INEQUALITIES
If m denotes the minimum value of the function f(x), we seek
the conditions under which
m ^ — l/k = —ajb,
i.e.,
a+mb ^ 0. C)
We shall show that m = /@). First of all, we use L'Hopital's
method to find
x cos x—sin x —a; sin a;
/@) = hm ; = lim
x-,o x—sin a; ^jl—cos a;
—sin x—x cos x . x sin x—2 cos x
= lrm ; = lim
B-.0 Sin X я_о COS X
__ 2
We must prove that f{x) ^ — 2, i.e., that
x cos x—sin x 2; —2ж+2 sin x {x S: 0),
жB+ cos ж) S2 3 sin ж (ж S2 0),
3 sin Д7
Since
B + cosa;)cosa;4-sin2a; /1—cosaA2
г' ж) = 1-3— = > 0,
6 v ' B+cos xJ \2+cos x/ —
the function g(x) is increasing, so that
which is identical with the inequality D). Hence m = — 2, and
the condition C) may be written as a S: 2b. Thus, we have proved
the inequality
a-\-b cos x
(Solution by D. Djokovic.)
5.12 Prove Everitt's inequalities:

TRIGONOMETRIC FUNCTIONS 107
sin(9 2A— cos 0)
sin 0 2+cos 0
@ < в ^л).
sin б 4—4 cos б —б2
sin б cos 6+4—б2/3
__ > _ .
Note. Concerning these inequalities and their generalizations
see: The Mathematical Gazette, vol. 44, 1960, p. 52 — 54.
5.13 Prove the inequality
Va+Vx+Vc ^ 4V3,
where
A = tan /3 tan y+5, В = tan у tan oc+5, С = tan « tan /3+5,
a > 0, /S > 0, у > 0, «+/3+y = ^7т.
5.14 If 0 < x, у < nj2 and 0 < x-\-y < л/2, prove that
0 < tan x tan «/ < 1, tan(«+2/) > tan ж+tan «/.
5.15 Prove that
sin e+i sin 26 > 0 @<<Э<7г), A)
sin 0+^ sin 26+1 sin 36 > 0 @ < 6 < тг). B)
Proof. From the identity
sin 0+! sin 26 = sin 0(l + cos 6),
since
sin 0 > 0 and 1 + cos 0 > 0 @ < 0 < л),
A) follows immediately.
In order to prove B), we consider the identity
sin 0+J sin 20+|-sm30 = sin0+sm0cos0+^sin0 Dcos20—1)

108 ELEMENTARY INEQUALITIES
= ^ sin 6B4-3 cos (9+4 cos2 в)
s=|sin 6{(l+cos 6J+(l+cos 6L-3 cos2 в},
from which B) follows at once.
The inequalities A) and B) are particular cases of Jackson's
inequality (see §0.8).
5.16 Prove that
sin 04-sin 204-... +sinn6+%sin{n+l)d ^ 0 @ ^ в ^ n). A)
Proof. The left-hand side of A) may be written in the form
Since
n+1
2*шЩ. B)
=1 k=l
sin - 6 sin в
n 2 2
У sin Ы = —j @ < б < n),
*=i sm -s-6
the expression B) is equal to
/ и„ . n+1 n+l . n+2
/ sin - в sm в sin в sin
12 2 2 2
+ ■
2 \ sin |-(9 sin \6
sm в
1 2 / . и , . «4-2
sin
/ и И4-2 \
(sm-fl+sm —fl)
This expression is non-negative for all values of в e @, л).
Finally, A) is trivially true if 6 = 0 or ж.
5.17 Prove that
arc cos x ^ Vl-я2 (—
This assumes the principal-value branch of the inverse function.

TRIGONOMETRIC FUNCTIONS 109
Proof. For t ^ 0,
t ^ sin < = Vl —cos2t.
If x e [—1, +1], we may set t = arc cos x i.e., x = cos t. Conse-
Consequently,
arc cos x 2s Vl— a;2 (—1 ^ ж ^ +1).
5.18 Prove that
arc sin x < x-\-Vx @ < x ±s 1).
Proof. If we set x = sin t @ < t r£ n/2), the given inequality
becomes
<—sin * < Vsin t. A)
By Taylor's formula,
t3
sint = t cos 6t @ < в < I),
i.e.,
t—sin t = — cos Qt < — .
6 6
The inequality A) follows from the inequality
i.e.,
t6
<sint @<t^n/2), B)
which, follows from Jordan's inequality
It
-^sint @ <t ^tt/2),
because in the interval under consideration
— <-, i-e., <5<72/тг.
36 л

110 ELEMENTARY INEQUALITIES
5.19 Prove the inequality
i-20/я ^ cose ^я/2-е (о^е
Hint. Use Jordan's inequality (§ 0.9).
in \ I ( П \
5.20 tan ax < I ^ sin «* / I 2 cos я*) «
\fo=l // \k=l I
@ < ax < a2 < . . . < an < \n).
5.21 Prove the inequality
2(cosa—sin яJ ^ [cos(a+e)+sin(a+e)]2
+ [cos(a— e)+sin(a — в)]2 ^ 2(cos я+sin яJ
where 0 < a < я/2 and 6 is arbitrary.
Solution. Taking into account that
[cos(a+0)+sin(a+9)]2+[cos(a — 0)+sin(a-0)]2
= 2+sin 2(a + 0)+sin 2(а — в) = 2 + 2 sin 2a cos 20,
2(cos a—sin aJ = 2 — 2 sin 2a,
2(cos a+sin aJ = 2 + 2 sin 2a,
the given inequality reduces to
-1 ^ cos 20 ^ +1.
5.22 sinz+tan x > 2x @ < x < л/2).
5.23 1 ^cosz ^ 1— \x2.
5.24 tana; > x+^x3 @ < x < я/2).
5.25 arc tan a; < ж—-|a:3 @ < x ^ 1).
5.26 x—^x3 < arc tan a; (a: > 0).
5.27 —log cos x < \ sin a; tan a: @ < a: < я/2).
5.28 F—a)cos& ^ sin&—sin a ^ (b — a) cos a @ < a, &<я/2).
sin же
5.29 я< ^4 @<ж<1).
жAж)

5.30
5.31
5.32
GEOMETRIC INEQUALITIES
tan a tan b sin a sin b
111
<
я
b a
sin b ,
> —-— (a < b; 0 < a, b < fя .
b
b
d" /sin ж\
d" /1—cosz\
1
5.33 cos x-\-x sin ж > 1 @ < x 5S я/2).
5.34 tan ж < ж+ж3 @ < x ^ я/3).
5.35 -V3^
2+cos
§ 6. Geometric Inequalities
6.1 Show that if a, /3, у are the angles of a triangle Л5С, then
cos a+ cos /3+ cos у 5S 3/2.
Solution. Let |5C| = a, [C4| = 6, \AB\ = с Since
AB-BC == —cacos/3, 5С-СЛ = — abcosy, С A • AB = —6ccos a,
we find
0^ 1 1 = 3 — 2(cos a+cos/3+ cosy),
\ я & с /
whence
cos a + cos /3+ cos у ^ 3/2.
(Solution by D. С. В. Marsh.)
6.2 Show that if А, В, С are the angles of a triangle, then
1°: (sin^/2+sin5/2 + si
2°: cos A /2 cos 5/2 cos С/2
3°: tan2 Л/2+tan2 5/2+tan2 С/2 ^ 1.

112 ELEMENTARY INEQUALITIES
6.3 Let a, b, с be the sides of a triangle ABC of area S.
Prove the inequality
4SV3 ^ a2+b*+c2. A)
Solution. Let us suppose that A) is false, so that
4SV3 > a2 + b*+c*. B)
The inequality B) is equivalent to
L), C)
2&csina>L(a
V3
where a is the angle opposite the side a. By the law of cosines,
26c cos a = b2+c2-a2. D)
Squaring C) and D) and adding, we obtain the inequality
a2b2+b2c2+c2a2 > а*+Ь*+с*,
which is equivalent to
(я2-62J+(я2-с2J+F2-с2J < О,
which is false.
From this result it follows that the inequality A) is valid.
6.4 Let P be any point in the interior of the triangle А ВС and
let \PA\ = x, \PB\ = y, \PC\ = z. Let p, q, r be the distances of
P from the sides ВС, С А, АВ, respectively.
Prove the inequality
xyz ^ (q+r)(r+p){q+p).
6.5 With the notation of 6.4, prove that
x+y+z ^ 2(p+q+r).
Solution. If L, M, N are the feet of the perpendicular from
P to ВС, С А, АВ, respectively, we have
MN = (q2+r2+2qr cos a)*, MN = x sin a,

GEOMETRIC INEQUALITIES 113
where a, /3, у are the angles of the triangle. We now have, in turn,
<*, 0, у
cos Y~~r cos /3J]1//2/sin a
sin # sin
(Solution by L. J. Mordell.)
6.6 With the notation of 6.4, prove that
фх+qy+rz ^ 2(qr+rfi+fiq), A)
B)
Solution. Denote the sides of the triangle by a, b, с and its
area by A. If ha is the altitude from A,
а(х-\-ф) ^ aha = 2Л = a-p
whence
яж ^ &2+cr. C)
From C)
whence
This proves A). For B), we apply the inequality of the arith-
arithmetic and geometric means to C), obtaining
ax ^ 2(bcqrIl*
and two similar inequalities. Multiplying these yields B).
(Note that B) may also be deduced from the result of 6.4.)
(Problem and solution by A. Oppenheim.)
6.7 Show that a necessary and sufficient condition that line

114 ELEMENTARY INEQUALITIES
segments of lengths a, b, с can form a triangle is:
6.8 Show that if ax, a2, a3 are the sides of the triangle А ВС and
hx, h2, hz the corresponding altitudes, then
^.з з
V3 J ak ^ 2 2 К
Equality holds if and only if ABC is equilateral.
6.9 Show that if A, B,C are the angles of a triangle ABC, then
sin A /2 sin £/2 sin С/2 ^ 1/8.
6.10 Show that if я, 6, с are the sides of any triangle, then
(a+b—c)(b+c—a)(c+a—b) <abc.
6.11 If ha is the altitude to the side ВС of the triangle ABC
(\BC\ = a, \CA\ = b, \AB\ = c), prove that
= a + b+c\
2 J '
6.12 Show that if r is the radius of the inscribed circle of a
triangle and R the radius of the circumscribed circle, then
r 1
6.13 A given point is at distances л/2, 2 and VS—1 from the
vertices of a triangle. Find the maximum area of the triangle.
Solution. Label the given point P and the vertices А, В, С
with PA = 2, PB = л/2, PC = V3 — 1. Let ж denote the length
of the perpendicular from P to A B. Then the optimum configura-
configuration gives an area of
{х2I/2} (O^x^ s/2).
This area has a maximum of ^C+V3), at x = 1.

GEOMETRIC INEQUALITIES 115
(Solution by D. С. В. Marsh.)
6.14 Let Ak (k = 1, 2, 3, 4) be the vertices of a square and P
a point in the same plane. Prove the inequality
2 \АкР\ ^ A + Л/2) max \AlcP\+min\AkP\
fc=l к к
and determine the points P for which equality holds.
6.15 If Sk is the sum of all the perpendiculars from the center
to the sides of a regular polygon of k sides which is inscribed
in a circle of radius r, prove that
6.16 About an arbitrary circle there is circumscribed a regular
polygon of n sides each of length an, and within this circle is
inscribed a regular polygon of k sides each of length bk. Prove
the inequalities:
«n+i <bn (n = 5,6,. . .),
<*n+i >bn (n = 3, 4),
2«n+i < K + bn+1 {n = 9, 10, . . .),
2an+1>bn+bn+1 («=3,4,..., 8).
This exercise may be found in the journal: Fiziko-matematicesko
spisanie, Sofija, v. 2, 1959, p. 239.
6.17 Given a circle С of circumference U. Let pn be the peri-
perimeter of the regular w-gon inscribed in С and Pn the perimeter of
the circumscribed regular w-gon (n ^ 3). Prove that:
1°: The sequence {pn} is monotone increasing, and
the sequence {Pn} is monotone decreasing;
2°: $pn+$Pn>U.
Solution. 1°: The side of a regular inscribed w-gon measures
2r sin (ж/п) and the side of a regular circumscribed w-gon is
2r tan (л/п), where r is the radius of the circle C. Therefore
fn = 2rn sinfa/n) = 2nr(nl7c)sin (ж/п), A)

116 ELEMENTARY INEQUALITIES
Pn = 2rn tan (я/и) = 2nr{njn) tan (я/и). B)
For 0 < x < я/2, the function p(x) = is decreasing and
x
tan ж
the function P(x) = is increasing, because
ОС
x cos x — sin x cos x
p'{x) = = —— (x—tan x) < 0 @ < x < я/2),
ОС ОС
X
—tan x
cos 2x x—sin x cos x 2x —sin 2x
w ж2 ж2 cos 2x 2 ж2 cos2 x
@ <x < я/2).
Since 0 < n\n < я/2 (и = 3, 4, . . .), it follows that
fin = 2лгр{л/п) \ , Pn = 2nr P{njn) ф (и = 3, 4, . . .).
2°: The inequality which we wish to prove may be written, by
means of A) and B), in the form
2r
— n
3
B sin- +tan-| > 2яг In = 3, 4, . . .),
\ n n/
or
2sin-+tan-> 3- (w = 3, 4, ...). C)
n n n
Let us set f(x) = 2 sin x-\-ta.nx — 3x. Since
/@) = 0, f'(x) = 2 cos ж+sec2 ж-3, /'@) = 0,
2 sin ж 2 sin ж ,
fix) = -2 sin жН = (l-cos3a; > 0
cos3a; cos3a;
@ < x < -|я),
the function f(x) is strictly monotone increasing on the interval
[0,я/2)Де.,
2 sin x+tan x> 3x @ < x < я/2). D)

GEOMETRIC INEQUALITIES 117
The inequality C) follows immediately from D).
(Solution by D. Adamovic.)
6.18 Given a tetrahedron with vertices A, B, C, D, let P be
any point in this tetrahedron. Let А', В', С, D' be the orthogonal
projections of the points A, B, C, D onto the faces BCD, CD A
DAB, ABC. Prove the relation
2 \PA\ (area ABCD) ^ 3^\PA'\ (area ABCD),
where, for example,
2\PA\ (area ABCD) = |РЛ|(агеа Д£С£>) + |Р£|(агеа ACDA)
+ |PC|(area Д£>Л£) + |Р.О|(агеа &ABC).
Hint. Let HA, HB, Hc, HD be the altitudes of the tetrahedron
dropped to the triangular bases BCD, CD A, DAB, ABC. Then,
HA ^ \PA\ + \PA'\,HB£ \PB\ + \PB'\, etc.
6.19 Let P be any point in the interior of an arbitrary tetra-
tetrahedron. Let -pv ft2, -p3, -pi be the distances from this point to the
vertices of the tetrahedron, and let qlt q2, qz, qt be the distances
from this point to the faces of the tetrahedron. Prove the in-
inequality
(This problem was contributed by W. Gridasow.)
6.20 Let a, b, с be the sides of a triangle; prove that
36
6.21 On the interval —1 ^ x :£ -\-l, let there be given n
distinct points Ръ P2, . . ., Pn. Let Tk be the product of the
distances of the point Pk from the other points. Prove that
n I
V > On-2
k=l I ic
6.22 1°: If a, b, с are the lengths of the sides of any triangle

118 ELEMENTARY INEQUALITIES
(which is nondegenerate), prove that
\a2 < &2+c2 ^ 2a2, A)
where a = max (я, b, c).
2°: Are the conditions A) sufficient for the existence of a
triangle (which is nondegenerate) with sides of lengths a, b, c?
Solution. 1°:
{b ^ а, с ^ a} => 62 + c2 ^ 2a2, B)
a < b+c => a2 < 62+26c+c2, C)
F-cJ ^ 0 => 26c ^ 62+c2. D)
From C) and D) it follows that
я2 < 2F2+c2) => |«2 < 62+c2. E)
From B) and E), we obtain A).
2°: If b = §я, с = \a, the conditions A) hold, but b+c = a,
which contradicts the supposition that the triangle is non-
degenerate.
Now, if b = \a, с = \a, the conditions A) are again satisfied.
However, with the lengths a, b, с so chosen, it is not possible to
construct a triangle, because b-\-c < a. Thus, the conditions A)
do not suffice.
6.23 Prove that if, with \BC\ = a, \CA\ = b, \AB\ = c, it is
possible to construct a triangle ABC, then with
\fr\ = Уа, И = УЪ, |oc/?| = Ус
(n a natural number > 1), one may construct a triangle a/Sy.
Proof. Without loss of generality, we may suppose that
0 <a^b ^c. A)
Since a, b, с are the lengths of the sides of the triangle,
с < a+b, B)
с > b-a. C)

GEOMETRIC INEQUALITIES 119
The relation A) for the triangle a/Sy corresponds to
0 < Ц~а < ЦЪ< П4с. D)
The problem may be solved if we can prove that the inequalities
B) and C) imply that
7c < Z/a+Z/b, and, E)
П1~ П/Г7 П1~ , .
sjc > sjb — sja. F)
Let us suppose that E) is false, so that
n4c ^ n4~a^b. G)
Then, after raising both sides to the wth power, we find
с ^ a + b+M, (8)
where M is a positive number.
Since (8) contradicts B) for all a and b, we have proved
that E) is valid.
Let us assume now that F) does not hold, but instead that
n/~ nnr ni—
Vе ^ Vo—Vя-
Then, after raising to the nth power, it follows that
a + c+M^b, (9)
where M is a positive number.
Since (9) contradicts C), we have proved the inequality F).
Consequently, if the inequalities B) and C) are valid, E) and
F) follow.
Remark. For the six relations
а-\-Ъ > с, Ъ-\-с > a, c-\-a > b,
a > \b — c\, b > \c — а], с > \a — b\,
which hold for the sides a, 6, с of a triangle, only three are independent,
the others following from these. If we use the assumption A), only a-\-b>c
needs to be checked. This means that in the proof it is possible to omit
the relations C) and F).

120 ELEMENTARY INEQUALITIES
§ 7. Inequalities Involving Mean Values and Symmetric
Functions
7.1 Given positive numbers a and b (> a), let
2ab
A = £(«+&), G = Vab, H =
a+b
Prove the following inequalities:
1°: A>G> H,
2°: A-G> G-H,
3°: A—G < (b-aJ/(8a),
4°: A-H < (b-aJlDa).
Proof. 1°: Since
A-G = \[V~a~~VbY > 0, A)
we have A > G. Since
H-G= -^(Va-VbJ<0, B)
we have H <. G.
2°: From A) and B) it follows that
A-2G+H = ^?=# > 0 => Л-G > G-H.
2(a+b)
3°: From A) we obtain
(a-b)*
A-G =
2(Va+VbJ'
whence, by the assumption that a < b, it follows that
A-G< (a-bJ/(8a).
4°: In a similar manner, we have

SYMMETRIC FUNCTIONS 121
7.2 Prove the inequality
1 1 1 a + b+c
1 1 < (a, b, с distinct and positive).
а о с а? о3 с3
7.3 Let dry, a2, . . ■, an be positive numbers no two of which are
equal. Prove the inequalities
1 у i
n
у i 1 J_ у i у _L /s - у а \
k^xak s ' n—lk=1ak ^s — Ub \ ъ~у k) '
7.4 If alt a2, ...,an> 0 and Pk = (< +
n
then
7.5 Let the sum of four positive numbers be 4c and the sum
of their squares 8c2; show that none of these numbers can exceed
Solution. Since the arithmetic mean of the squares is not less
than the square of the arithmetic mean,
3
i.e.,
(Solution by D. Djokovic.)
7.6 Use Bernoulli's inequality,
A+в)»^1+я« (а>-1,п= 1,2, 3, ...), A)
to prove .
b-l
(Ъ > 0; m= 1, 2, 3, . . .), and B)
(e > o, i > 0). C)

122 ELEMENTARY INEQUALITIES
Proof. From A) we have
/ b—l\m b — 1
H ^ 1+m = 6, D)
\ m ] m
if (b-l)lm > — 1 and m = 1, 2, 3, . . ., i.e., if b > \—m and
m = 1, 2, 3, .... The inequality B) follows at once.
To prove C), we replace b in B) by bja to obtain
m
Multiplying both sides of E) by a yields
b — a (m— l)a-\-b
^ a\=
m m
which is C).
7.7 Prove the inequality
3(as+bs + cs) ^ (a2+b2+c2)(a+b + c) (a, b, c> 0)
and from this result deduce that
a b с 3
TV~ + -I- + -71 ^ T («, 6, с > 0).
6+c c + a a+6 2
7.8 Let bx, b2, . . ., bn be any permutation of the positive num-
numbers ax, a2, . . ., an; then show that
n
J,ajbk ^ n.
k=l
Solution. This result is an immediate consequence of the
inequality between the arithmetic and geometric means, since
7.9 Prove the inequality
pa+qb+rc ^ a'b'c? (a, b, c, p, q, r > 0; p+q+r = 1). A)

SYMMETRIC FUNCTIONS 123
Proof. Applying the inequality between the arithmetic and
geometric means, we obtain
I n in il/« / 1 n \
- 2 /(*») > П/Ы = exp - 2 log/(*„).
И y=l U=l i \П к=1 /
whence, taking limits as и —> со,
/•£ л»
/(z)dx ^ exp log/(ж)da;, B)
J a J a
where we assume that f(x) is positive and piecewise continuous
over the interval [a, /?].
In particular, if we set
(a @ <x <p),
f[x) = \b (p<x< p+q),
[c (p+q<x < 1),
in B) we obtain the inequality A).
(Proof by D. Djokovic.)
7.10 If p0 = 1, pk = — (k = 1, 2, . . .,n), where ak is the ele-
mentary symmetric function of order k of the positive numbers
ax, a2, . . ., an, then show that
Рг ^ PJ'2 ^ • • • ^ prW ^ . . . ^ pnV».
Remark. For this result see: J. W. Archbold; Algebra, London 1958,
p. 53 — 55.
7.11 Prove the inequality
b-\-c c-\-a a-\-b
Solution. From the inequalities
%(a+b+c) (a, b, с > 0). A)

124 ELEMENTARY INEQUALITIES
b-\-c c-\-a a-\-b
~Y ^ Vbc' "T" - Vca> ~J~ - Vab'
we obtain
b-\-c \/bc' c-\-a л/са a-\-b л/ab
It follows that
be ca ab
b-\-c c-\-a a-\-b ~
Using B) again, this inequality yields
be ca ab /b-\-c c-\-a a-\-b\
1 1 < A I 1 1 = \(а-\-Ь-\-с).
b + c c+a a + b ~ 2 \ 2 2 2/
Equality holds in A) only when a = b = c.
7.12 Prove the inequality
%(am+bm) < {i(a + b)}m @ < m < 1; a, b > 0; а ф Ъ).
Generalize.
7.13 Prove that if
П
Sr = 2 ak and fir = 2 «1«2 • • • ar
(i.e.,/>risther-th elementary symmetric function of ax, a2,.. ■ ,an),
then
^ (m 1 M
(a1( a2, . . .,а„ > О).
sr r\(n—r)l
7.14 If a, b, с are three distinct real numbers, prove that
3 гшп(й, b, c) < 2 a~ B д2~~2 a^I/2
< 2 a+ B д2 — 2 a^)x'2 < 3 тах(й, 6, с),
where

SYMMETRIC FUNCTIONS 125
2 a = a + b+c, J a2 = й2 + 62+с2, J яб = й6+6с+сй.
Proof. The function
f(x) = (ж—a)(x—b)(x—c) = x3— B й)ж2+ B ab)x—abc
vanishes for x = a, x = b, x = с We deduce that f'{x) vanishes
for values of x which are solutions of
Зж2-2B а)х+2,аЬ = 0.
These solutions lie between min (a, b, c) and max (a, b, c). From
this we obtain the proposed inequalities.
7.15 HAk and Gk are the arithmetic and geometric means of the
first k numbers in the sequence ax, a2, ■ ■ ■ {ai > 0), prove that
k(Ak-Gk) ^ (*-l)DH-CH).
(See the paper of L. Tchakaloff: Sur quelques inegalites entre la
moyenne arithmetique et la moyenne geometrique, Publications de
I'Institut mathimatique, Beograd, v. 3A7I963, p. 43).
7.16 Let ak (k = 1, 2, . . ., n) be the elementary symmetric
functions of order k of the real numbers xx, x2, . . ., xn. Prove
that
{ax > 0, a2 > 0, . . ., an > 0} о {xx > 0, x2 > 0, . . ., xn > 0}.
Example. If x, y, z are real, then
{x-\-y+z > 0, yz-\-zx-\-xy > 0, xyz > 0}o {x > 0,у > 0,z>0}.
Proof. We see at once that
xx > 0, x2 > 0, . . ., xn > 0 implies ax > 0, a2 > 0, . . ., an > 0.
In order to prove that the converse holds, we form the poly-
polynomial equation with roots xx, x2, . . ., xn, namely,
xn—a1xn~1+a2xn-2— . . . +(-1)истп = 0.

126 ELEMENTARY INEQUALITIES
We determine whether this equation may have negative roots
by replacing x by —x, whence
xn+a1xn~1+a2xn-2+ . . . +an = 0.
If the conditions аг > 0, cr2 > 0, . . ., an > 0 hold, the poly-
polynomial
xn+a1xn-1+a2xn~2+ . . . +an
does not vanish for any positive value of x. Thus, since this
polynomial vanishes for x = —xk(k = 1, 2, . . ., n), it follows
that
xk > 0 (k = 1, 2, . . ., n).
Thus
ax > 0, a2 > 0, . . ., an > 0 implies xx > 0, x2 > 0, . . ., xn > 0.
7.17 1°: Does
(A) x > 0, у > 0, z > 0
imply
(B) x+y+z > 0, xyz > 0?
2°: Does (B) imply (A)?
3°: Are (A) and (B) equivalent?
7.18 Prove the inequality
n
2 xixk 5: 0 (x1, x2, . . ., xn real).
i,k=l
Proof. This inequality follows directly from the identity
n / n \ 2 n
" Z. лгхк — \ Z, xk\ \ Z,xk ■
7.19 Prove the inequality {а-\-Ъ-\-с){Ъс-\-са-\-аЪ) > 9 abc, where
a, b, с are positive numbers which are not all equal.
Proof. (a-\-b-\-c)(bc-\-ca-\-ab)—9 abc

SYMMETRIC FUNCTIONS 127
= a(b2+c2)+b(c2+a2)+c(a2 + b2) — 6 abc
= a(b-cJ+b{c-aJ+c(a-bJ > 0.
Remark. What changes must be made in the given inequality
if a, b, с are negative numbers?
(»—1 \ n n—1 к
2 (KI ^
2 (»-
k=0
7.21 fxk^n Lk>0,
7.22 (b + c)(c+a)(a+b) ^ 8 a5c (a, b,c^ 0).
7.23 6сF+с)+сй(с+й) + й6(й + 6) ^ 6«6c (a.b.c^O).
7.24 (— — lW 1) ( 1) ^ 8 {a, b, с > 0; a+b+c=l).
\a } \b / \c I
7.25 b2c2+c2a2+a2b2 ^ )
7.26 (a2 + b2+c2+d2+e*)(as + b3+cs+d3 + es) ^ 25 abcde
(a, b, c,d,e^ 0).
7.27 as + bs + cs+15abc ^ 2(й+5+с)(й2 + 62+с2) (а, й, с ^ 0).
7.28 (s—e)(s — b)(s—c)(s—d) ^ 81a6crf {a, b,c,d> 0;
s = a-\-b+c-{-d).
7.29 -^_ + __+-f- + —-^2 (e,6,c,rf>0).
6+c c+d d+a a+b
7.30 л/^+л/^+л/^+л/^+л/^+л/^ ^ |(a+6+c+^)
(a, b,c,d^ 0).
и s n2 / n \
7.31 J > 7 K>0;s=2«* •
и Й fl ( n \
7.32 T—^> \ak > 0;s = T ak .
7.33 f

128 ELEMENTARY INEQUALITIES
7.34 a+b+c ^ apbqcr+arbpc'i+aqbrc'p
(a,b,c,p,q,r> O;p+q+r= 1)
7.35 Bn+l)xn ^ l+x+x2+ . . . +x2n (x ^ 0).
7.36 l~x2n ^2nxn(l-x) @^ж^1).
7.37 (abc)&(as+bs+cs)& ^
7.38 -I«/ ^-id (o <P<я;ч ^ o).
7.39 п -2 «*"i ^-2«" »»= 2a^ a*> «* > о .
<=1 \И j=l / »,=l \ i=l /
(a, b,c ^ 0).
7.40 2(я-],_ _
(х1, х2, . . ., хп real numbers).
а Ъ с
7.41 #
Ъ-\-с с-\-а а-\-Ъ
(where а, Ъ, с are the sides of a triangle).
7.42
n , n \
7.43 2 afcm = w w > x» a* > °> 2 au = n
k=l \ k=l '
7.44 (n-
r, s=l
Г<8

CAUCHY-SCHWARZ-BUNIAKOWSKI
129
§ 8. The Inequalities of Cauchy-Schwarz-Buniakowski,
Holder, Minkowski and Chebychev.
8.1 (йос + 6/?J ^ (й2 + 62)(ос2+/?2).
8.2 (a4 + b2c+c2a)(ab2+bc2+ca2) ^ ЫЧ2с2 (a, b, с ^ 0).
8-3 (i)(i
,2 я я J 2
4 (Tab)
•5 2a* ^ 2 ** + I 2 ak ) («;
(n \ 4 n n / n
^ к к kj = £* к £* к I Z*
8
8
8.6
8.7
8
(n a \2 n
«A
2 n
= 2 la*l2 2 №к\2 (ак> bk complex numbers).
^ i B l«*l = 2 \bk? = H
(aft) 6ft complex numbers).
8.9
8.10
8.11
8.12 2 ^"
8.13 К + б^^йа+баK ^ «/a29+
8.14
; я2,
0).
^ 1+жг2/1-г [х, у ^ 0; 0 <г < 1).

130 ELEMENTARY INEQUALITIES
8.15 [{a1+b1)r+ {a2+b2YY'r ^ (a1'+a2'Il'+ (V
8.16 П (l+ak) ^ A+6)" L > 0; П ak = Й .
8.17 ПA+а4)^2- (ak>0;Uah=l).
(a, b, ol, p ^ 0).
8.18 [{a+b) (я
8.19 [(a1+b1)
8.20 [(a1+b1) (a2+b2) (as+b3) K
К 6, ^ 0; k = 1, 2, 3).
(ah, bh ^0;6= 1,2,3,4).
8.21
2 n \
2
8.22 ( J akbkcX ^ 2 %4 2 V 2 b^ J V J c*4 2 ^2-
\fc=l / fc=l k=\ k=l k=l k=l k=l
i £ /+e ^ (- 2 «/) (- 2 «t«) (/>, ? > 0; «t 2> 0).
8.23 i
/ 1 « \ * 1 n
8.24 -2M ^ - 2 */ К ^ 0).
\n v=1 I n v=1
(n \ 2 n
8.26 bc(b+c)+ca(c+a)+ab(a+b)^2(a3+b3+c3) (a.b.c^O).
8.27 Fс+ся+я6)(я+6+сL ^ 27(я3 + 63+с3J (я, 6, с ^ 0).
8.28 (я4+64+с*+^4)(я3+63+с3+^) ^ 4(я'+6'+с7+^7)
(я, b,c,d^ 0).
8.29 (в+6)(«2+&2) (я3+63) ^ 4(я6+66) (я, 6^0).
8.30 (я+6)(я3+63)(я'+67) ^4(я11+6") (а, 6^0).

INTEGRALS 13l
8.31 (a2+b2){a3+bs)(a6+b6) ;g 4(я"+611) (a, b ^ 0).
8.32 ab{a2+b2) ;g я4+64.
8.33 a2b2{a5+b5) ^ a9+b9 (a, b ^ 0).
8.34 ^(l+a;™-1) ^ 2 Ж ~ (ж > 0; и ^ 2).
Л/ 1
8.35 (я+6)™ < 2™-!(й:™+&я) (и ^2; я, 6 > 0; а фЬ).
8.36 лп—1 ^и(я<"+1>/2—я»"-1»/2) (л ^ 1).
8.37 -^ + —=^ \/«+V& (л,6>0).
8.38 (я4+64)(я5+65) < 2(я9+69) (я, Ъ > 0; я ^ Ь).
8.39 (я3+63)(я2+62) < 2(я5+б5) (а,Ь> 0;а ф Ъ).
§ 9. Inequalities Involving Integrals.
9.1 Let
In = 2n cot™ я tan™ x dx l0<«< — ;иа natural number I .
Prove the inequality
tan™ x dx <M tan" x sec2 a; da;, A)
Jo Jo
and hence deduce that
In < 2 tan л. B.)
Will the inequalities A) and B) be valid if n is an arbitrary
real number?
9.2 Prove the inequality
q{aP-l) ^ p(x"-l) (j> > q > 0; x > 0)
and from this result, by integration, prove the inequality
1 ( xv } 1 ( x"
— { 1 > — {
)
— 1 (и a natural number)..

132 ELEMENTARY INEQUALITIES
9.3 Let
In = Г tan" xdx (n = 2, 3 . . .; 0 < a < %n).
Jo
Prove the relation
In+h-ar
n— 1
Compute I3 and 74) and thereby deduce that
a-\-\og sec a < tan а+\ tan2 a—\ tan3 a @ < я < тг/4).
9.4 Prove Esseen's inequality
ф(-а-Ъ) ^2ф{-а)ф{-Ъ) (а.Ь^О), A)
where
Proof. Consider the function
F(a, b) = ф{-а-ЪIф{-а) {a, b ^ 0). C)
Differentiating we obtain
^ = p^a) {ф'(-а)ф(-а-Ь)-ф'(-а-Ь)ф(-а)}
e-a2/2
2(—а)
where we have set
G(a, Ъ) = ф(-а-Ь)-в-Ьь2-аЬ ф{—а) (a, b ^ 0). E)
From this result we find
_= Ъе-^~аъ ф(-а) ^0,
да
whence
G(a, b) ^limG(«, b) = 0,
a->+co

INTEGRALS 133
and thus D) gives
8F <
Accordingly,
F{a, b) ^ F@, b) = ~
which completes the proof of Esseen's inequality.
(Proof by D. Djokovic.)
Remark. Starting with Esseen's inequality, prove that for the function
Vn Jo
we have the analogous inequality
f(a)f(b) S f(a)+f(b)-f(a + b) (a, b S 0).
9.5 Let f(x) be a continuous, non-negative function for all
real x, and suppose that
f(x)dx = 1.
J —CO
If
г
=
J
+оо
f(x)cosxt dx (t real),
prove the inequality
gBt)
Proof. Since
cos 2xt = 2 cos^xt—l,
Л+ОО
gBt) = 2 f{x) co&xt&x— 1,
J —oo
whence the given inequality may be written in the form
/(a;)da; f(x)cos2xt dx > J /(a;
J —CO J —CO I J —CO

134 ELEMENTARY INEQUALITIES
Since the functions Vf(x) and Vf(x) cos xt (t ф 0) are linearly
independent, we see that the preceding inequality follows from
the Cauchy-Schwarz-Buniakowski inequality
Г tf(x)dx Г f2*(x)dx > ( Г fx{x) f2(x) dxY,
J a J а У J a )
where fx{x) and /2(ж) are two continuous, linearly independent
functions, when we set
a = —oo, Ъ = +oo, f^x) = Vf(x),f2{x) = Vf(x)cosxt (t фО).
(Solution by D. Djokovic.)
9.6 Prove the inequality
— > I sinn0d0| >— (n a natural number).
2n \Jo / 2(я+1) V ;
9.7 Let the function f(x) be positive and non-increasing over
the interval [1, -f oo). Show that if
gn(t) = tn f(tn) (t> l;n = 0,1,2, ...),
then
9.8 Prove the inequality
Bn)!!
Proof. Start with the following pair of inequalities:
Л+°о _ Л1 /*V™
e-^2 дх^л/nl e~nx% da; = е-ж2 da; {n = 1, 2,. . .); B)
Jo Jo Jo
1-х2 ^ e-*\ C)
From C) there follow in succession
A-a;2)" <^e-m* {n = 1, 2, . . .), D)
Г (l-a;2)nda; ^ f e-^dx. E)
Jo Jo

INTEGRALS 135
From B) and E) we find
f+oc 2 _ f1 _ Bn)\\
e~x dx ^ л/п I A — x2)n dx = л/п ~—'-—-.
Jo J о \^w-\-1)!!
9.9 Prove the inequality
Jo ' -V*^=2)M2" («-1'2'"-)- A)
Proof. Starting with the inequality
e-^^-J_. (x ^ 0),
we obtain
l
- (JU ^- \J fit 1 it, . . . I
+00 . Л+00 1 Г'2 Bи — 3)П л
e~nx* dx < da; cos2"-^ d^ ^'
Г+00 . Л+00 1 Г
e~nx* dx < , da; =
h ~Jo (l+^2)n Jo
s^ d^ .
Bn-2)!!2
If we set хл/п = y, the preceding inequality yields A) at once.
9.10 Verify that
and from this result deduce the inequality
fVl—x
-п{л/2—\) <
Jo 1+х*
л/l-a
9.11 If fo(x) > 0(x ^ 0) and fn(x) = ^/^(Od/ (n = 1, 2,
prove the inequality
.« /v.fi+1
X X
9.12 Prove the inequality
«+OO ^
e~x*dx <—e- (a > 0).
Jo «
A)

136 ELEMENTARY INEQUALITIES
Proof. If x > a > 0, then
e-x* < e~ax, since x2 > ax.
Since
ль ль
e~x dx < e~axdx (b > a> 0),
J a J a
one has
lim e-x dx < lim e-M dx (a > 0),
Ь -> +oo J а Ь -> +oo J a
whence the inequality A) follows.
9.13 Prove the inequality
Bл)-У2 Г" e~x%l*
J —a
9.14 Prove the inequality
f L
Jo A — 2x cos
dx^
l — 2xcost+x2K/2 ~ 2t2
Hint. Start with the equality
dx = da;
Jo (l-2xcost+x2K!2 Jo [sin2if+(a;-cosifJ]3/2
Remark. It is also true that
141 1 . 9л2
9.15 If

INTEGRALS 137
prove the inequality
as well as the sharper inequality
л л 1
I
2 2 (I я2I/2
9.16 Let f(t) be a non-negative function defined on the interval
[0,1], and let
f(h)+f{h) /t-i+t2\
Prove the inequalities:
f /@ d/ ^ 3" Г tnf(t)dt (n = 0, 1, 2, . . .),
Jo Jo
2 f1 f1 2 f1
, f(t)dt < tnf(t)dt< f{t)dt.
(n+l)(n+2)]0n> -Jo /u -n+2j0/U
These inequalities were stated and proved by B. Sendov and
D. Skordev in the journal: Fiziko-matematicesko spisanie, Sofija,
v.2, 1959, p.240 and v. 3, 1960, p. 55.
9.17 Prove the inequality
f \f(x)-g{x)\2dx^2C\f(x)-h(x)\*dx+2C\g(x)-h(x)\*dx.
J a J a J a
(i)
Hint. f(x)-g(x) = {f(x)-h(x)}-{g(x)-h(x)}t
\f(x)-g(x)\ £ \f(x)-h{x)\ + \g(x)-h(x)\,
\f(x)-g(x)\* ^ |/(a;)-A(a;)
Since 2\f(x)-h{x)\ \g(x)-h(x)\ £ \f(x)-h(x)\*+\g(x)-h(x)\*,
the preceding inequality becomes
Remark. What conditions must f(x), g(x), h(x) satisfy, in order
that the inequality A) is meaningful?

138 ELEMENTARY INEQUALITIES
9.18 Prove the inequality
1.462 < Г <?%dx < 1.463.
9.19 Starting with Jordan's inequality
n
prove the inequality
/чг/2
f
n
2R
A)
Proof. For 0 < x < n/2, A) implies each of the following:
sin x > {2[n)x, — R sin x < - {2Jti)xR, e~Rslnx < e-ixRl",
Г
Jo
e~Rslnxdx
С
Jo
Я-/2
0
§ 10. Inequalities in the Complex Domain.
10.1 Prove the inequality
Ь \a-b\ ^ \a\ + \b\,
where a and b are complex numbers.
Proof. 2|л| = \(a+b) + {a-b)\ ^ |л+6Ц-|л-6|,
2|6| = \(a+b)-(a-b)\ ^ |a+6| + |a-6|;
adding and dividing by 2 gives the stated inequality.
10.2 Show that the inequalities
Re z < 1/2 and
\—z
< 1
are equivalent.

THE COMPLEX DOMAIN
139
10.3 Show that the inequalities
1
\—z
are equivalent.
10.4 Show that the relations
Re < 1/2 and \z\ > 1
z—a
1—dz
< 1,
z—a
1—dz
*• у
z—a
1—dz
< 1)
are equivalent, respectively, to
\z\ < 1, \z\ = 1, \z\ > 1.
10.5 Prove the inequality
\k=l I \k=l fc=l
where the ak are real and the zk complex numbers.
Remark. This inequality is stronger than Schwarz's inequality
2<
k=l
k=l
*;=1
since
k=l
^ 2
k=l
10.6 Prove the inequality
|log(l+*)| ^ log(l + H) (z = x + iy),
where log z is the principal value of the logarithm.
10.7 Prove Kloosterman's inequality
kl2
\ez— (l-\-z/n)n\ < —M (z = x+iy Ф 0; n a natural number).
Remark. See: Wiskundige opgaven met de oplossingen, Achttiende
Deel, eerste stuk, 1943, p. 70 — 72.

140 ELEMENTARY INEQUALITIES
10.8 Prove or disprove the inequality
|z»+l_l| > \z\n |г_!| (г = х + {у; Re Z ^ 1). A)
Proof. Note that we must have z Ф 1. Let r = \z\, 6 — arg 2.
By reasons of symmetry, we may take 0 ^ 0 < тг/2. Since
|^|«+1—1 > |г|п+1_|г|п (y > !)(
the inequality A) holds if 2л—{п+1)в ^ 9, i.e., if
0£
B)
For и = 1 and n = 2, the condition B) is satisfied for the
entire region D = {z\Re z ^ 1, z Ф 1}, and for these cases the
inequality A) has been proved.
If n > 2, the inequality A) has been proved for the region
E = {z\Rez ^ 1, z ф 1, |0| ^ 2л/{п+2)}. In polar coordinates,
A) reads
2 r"^1 cos 0+1 > r"!+2rn+1cos(ra+lH (rcosd^l). {!')
This relation certainly holds if
2r2n+l > r2n + 2rn+1o/(r) = r%n — 2гп+г + 1 >0. C)
Since the polynomial f(r) has only two positive zeros: r — 1 and
r = rt > 1, C) holds for r > yx.
Since rx > 1 and 2 rx2 > l+^i, we have
)-(r12-l) = 0,
V(»'i-l)(»'i"-8+''i"^+ . . . + IJ =
whence
Consequently, we have proved that the inequality A) holds for
the region F = {z|Re z Si 1, r ^ r0}. Moreover, we may show
that E и F = D. This relation is equivalent to the inequality
^ D)

THE COMPLEX DOMAIN
141
y\
о
Fig. 8.
where a is the angle shown in Fig. 8. Thus
whence
l/ 1
sin a = I/ 1 =
' 2
2 2
sin a < = <
E)
Applying Jordan's inequality sin в > 2вIn, we obtain
n n
a < - sin a <
2 и—
from which we deduce that D) holds for n = 4, 5, 6, . . ..
For и == 3, we find from E) that sin a = л/о/З, so that
a < 49° and so D) is satisfied. This completes the proof.
(Solution by D. Djokovic.)

142
ELEMENTARY INEQUALITIES
10.9 Prove the inequality
\a+b\
\a\ \b\
A)
Proof. Let us assume that A) does not hold, but rather that
!«+*! l«l 1*1 B)
l-)-|a-)-o| 1 + |я| JL —I— jo |
Then we find that
\a+b\ > \a\ + \b\ + 2\ab\ + \ab\ \a+b\, C)
which is impossible because \a-\-b\ Ss \a\
Remark. Is the generalization
n
1 +
S ak
*=i
valid?
10.10 Prove that
\z—a\ \
10.11 If \a\ < 1 prove the inequality
0.
z—a
< exp
/2A-H)\
\ |1/г—a|2/ "
1—dz
Proof. Making use of
and
log(l+a;) ^ x {x ^ 0)
we find for |z| ^ 1, \a\ < 1
г—a
log
l — az
\l—az
1 — az\2
A)
<2
B)

THE COMPLEX DOMAIN
143
Hence, A) holds for \z\ ^ 1, \a\ < 1. On the other hand, we
evidently have
C)
-^-$7 < 1 for |*| < 1, \a\ < 1.
|l-ez|
By B) and C) the given inequality is established.
10.12 Prove that
1 "
— 2 *»
n „=i
v=l
where zlt z2, . . ., zn are arbitrary complex numbers.
Hint. Use the identity
i
n ,=i
where
n
1 "
w = — X zv.
П v=l
10.13 Let zlt z2, . . ., zn be complex numbers such that
а—в < arg zv < a+e @ < в < л/2).
Prove the inequality
2*.
S2 COS в ^ \z»\-
Solution. We have
v=l
Re {e-
™ i zp\
= 2 \zv\ cos(—a
Hence, the inequality is true.
^ cos

144 ELEMENTARY INEQUALITIES
§ 11. Miscellaneous Inequalities.
11.1 Prove that the following proposition is valid:
P(n):/(n) = V
Proof. P(l) is valid, because certainly 1 < 2. Suppose now
that P(n) is valid. Then,
/(n+1) = Vn+l+f(n)
(by the induction hypothesis)
<
Thus, P(n) => Р(и+1).
This completes the inductive proof.
11.2 Prove the relation
(са'-асУ < ЦаЬ'-а'ЩЪс'-Ъ'с)
=> (б2—ас > 0 and b'2—a'c' > 0).
11.3 Determine the values of n for which the following inequality
is valid
/n (Ж11X2> • • ■ > xn) — ; I : г • • • H : I ; = ~z
+ + + 2
(xt ^ 0; xt+xi+1 > 0, i = 1, 2, . . ., n; xn+1 = хг). A)
Proof for n = 3. For this case A) reads
+ + ^ (ж+ж «
= e8)- B)
We have
/*i , *2 , жз\ / , _|_ ч

MISCELLANEOUS INEQUALITIES 145
-? + - +ад М + ^ +хях1 р + -?
Accordingly, it is sufficient to prove the inequality
(а;1+а;2+ЖзJ ^ f (a^ + a^+a^) = 3(я:1а;2+а;2а;3+а;3а;1).
Since the latter may be written in the form
we conclude that it is valid. Equality holds in B) only if xx =
x% x$.
The inequality A) may be similarly proved for n = 4, 5, 6.
(See 7.7 in reference to the case n = 3.)
Remark. This problem was proposed by H. S.j Shapiro (The American
Mathematical Monthly, v. 61, 1954, p. 571). He proved A) for n — 3,
4 (unpublished).
However, the proof of this inequality for n = 3 predates Shapiro's
proposal. See, for example, G. L. Neviazhskii, Inequalities, Moscow 1947,
pp. 130—131.
The challenge of the inequality A) has created lively interest among
mathematicians, as may be seen from the historical sketch which follows.
That A) is not valid for all n was shown, by means of a counter-
counterexample, by Lighthill [The Mathematical Gazette, vol. 40, 1956, p. 266).
Here is his counter-example:
Let n = 20, x№ = ake, ж2*_г = l+bke (fi = 1, 2, . . ., 10), where
ax = 6, a2 = 5, a3 = 4, a4 = 3, ab = 2, a, = 1, a, = 2, a8 = 3, a9 = 4, aw = 5,
bj = 5, 62 = 4, 63 = 3, &4 = 2, 65 = 1, &6 = 2, &, = 3, 68 = 4, &9 = 5, b10 = 6.
and e is sufficiently small and positive.
We have
l + {ak+bk+1)e

146 ELEMENTARY INEQUALITIES
i . , ^к ^
= ake-ak(ak+1 + bk+1)e2 + O(e3) (Й = 1, 2, . . ., 10),
where we have set
ж21 = хъ ж22 = ж,, я„ = «!, Ьц = &!•
Summing these twenty equalities we find
f20{x1} ж2 , ж20) =
The coefficients p and 5 are:
10 10
p = S (Ь*-а*-Ь*+1)+ S a* = 0,
10 10
4 = S (a*+b*+x) (as+b*+i—*>*) — S
k=l it=l
and so the above relation becomes
/20(ж1; ж2, . . ., ж20) = 10-e
Thus, for sufficiently small e, ^„(ж!, ж2, . . ., ж20) < 10, which means
that for n = 20 the inequality A) is false.
Many writers have proved the truth of A) for n = 3, 4, 5, 6 or for some
of these values. A short proof of all these cases was given by L. J. Mordell
(Abhandlungen aus dem Mathematischen Seminar def Universitat Hamburg,
Band 22, Heft 3/4, 1958, p. 229 — 240). The proof given here for the case
n = 3 was taken from this source. These results of Mordell suggest the
conjecture that A) is false for all я S: 7.
In addition to the aforementioned article by Mordell, there is the proof
by A. Zulauf that A) does not hold for even n >. 14. Zulauf proved that
A) fails for n = 14 by using Lighthill's method, and then applied induc-
induction.
This method of proof cannot be applied to the cases n = 8, 10, 12,
because the coefficient q is then a positive-definite quadratic form in the
variables ak and bk. On this point see: Dina Gladis S. Thomas (The Ameri-
American Mathematical Monthly, vol. 68, 1961, p. 472 — 473).
Here is Zulauf's proof: Analogous to the counter-example quoted above,
we find that for n = 14 and sufficiently small e,
le, l + 4e, be, 1+e, be, 1, 2e, 1+e, 0, l + 4e, e, l + 6e, 4e)
< 7.
We suppose that A) does not hold for n = N and let xk(k = 1, 2, . . ., N)
be numbers such that
fN(xlt xit..., xN) < %N-

MISCELLANEOUS INEQUALITIES 147
Then
/fftife. жг, . . ., xN, xN_lt xN) = fN(xt, хг, . . ., xN) + l
i.e., A) fails for n = N-j-2 also. Since A) does not hold for n = 14, it
follows that A) fails to hold for all even n 5: 14.
R. A. Rankin (An Inequality, The Mathematical Gazette, vol. 42,
1958, p. 39 — 40) proved the relation
where ц(п) is the lower bound of the function /„(x1( хг, . . ., xn) for xk Si 0.
Thus it follows that A) does not hold for all sufficiently large n.
In a second article (The Mathematical Gazette, 1959, vol. 43, p. 182 — 184)
A. Zulauf proved that A) does not hold for n = 53, whence it follows that
A) is false for odd n 2: 53.
D. Z- Djokovic has proved (Proceedings of the Glasgow Mathematical
Association, vol. 6, part 1, 1963, p. 1—10) that A) is true for n = 8.
On the basis of this result P. H. Diananda (Proceedings of the Glasgow
Mathematical Association, vol. 6, part 1, 1963, p. 11 —13) and B. Bajsanski
(Publikacije Elektrotehnickog fakulteta Univerziteta и Beogradu, serija:
Matematika i fizika, No. 76, 1962) have proved independently of each
other that A) is also true for n = 7. In the same paper P. H. Diananda
has proved that A) does not hold for n = 27.
Hence, there is still undecided question whether A) is true or not for
n = 9, 10, 11, 12, 13, 15, 17, 19, 21, 23 and 25.
Equally interesting generalizations of this inequality are given by
P. H. Diananda (The American Mathematical Monthly, vol. 66, 1959,
p. 489-491).
11.4 Let
+ ^+ .+ , ,
Х<£-\-Х§ Хп_1-\-Хп Xn-\-X1
where хг, хг,..., х„ ^ 0; x1Jrx2, x2-{-x3,..., xn_1+xn, xn+x1 > 0.
Show that
1 < Q(x1,x2, . . .,xn) < n—1.
Remark. This result was proved by A. Zulauf (The Mathematical
Gazette, vol. 43, 1959, p. 42).
11.5 Determine whether the inequality
0 ^ (x+a)*l{x*+x+l) £ f («2-
holds for all a and x.

148 ELEMENTARY INEQUALITIES
11.6 If x = 2t/(l+t2), у = A-*:2)/A+*:2), prove that
2 ~9—8x—Ъу =
11.7 If alt a2, . . . are distinct positive numbers, prove that
and that in general, for 2 jg r,
k=l \k=l
11.8 Prove the inequality
P Р+Я Я
ft-\-m ft-\-q-\-m-{-n q-\-n
/m n . , , \
I — > — ; m, n natural numbers; ft, q > 0 .
\p q I
Deduce that
Aft+r &q+s Aft+q+t
In general, if f(x) is a monotonic increasing function, show that
11.9 Prove that
xn
(x > 0).
. . . +x2n ~ 2n+l
Proof. Making use of
t+t-1 ^ 2 (t = x", v = 1, 2, . . ., »; x > 0)
we obtain
ж" 1 1
2и
(Solution by R. Lucid.)

MISCELLANEOUS INEQALITIES 149
11.10 Prove the inequality
1+а+а?+ . . . +an n+1
• •> (a > 0; n a natural number > 1).
a+a2+a3 + . . . +an~1 n — 1 '
A)
Proof. For n = 2, A) is valid. Suppose that it holds for n = k,
i.e., that
l+a+a*+ +ak k+l
a+a2+a3+ . . .+ a*'1 = k— 1
Since a > 0, B) may be written in the form
k+1
whence we have
k+1
\ + a+ . . . +ак+ак+1 ^ —— (a+a2+
We shall establish
k+1 k+2
C)
« — 1
D)
proving that A) is also valid for n = k+1.
Let us assume that D) does not hold, but instead that
k+1 k+2
Ц (+*++ак-1)+ак+1 < -±- (a+a2+ ...
k
k-1
Then we find that
2(а+аг+ . . . + ak~1 + ah) + (ki+k)ah(a— 1) < 0,
which is clearly impossible if a S: 1. Thus, by induction, A) is
established if a Si 1.
If we set a = 1/6 (b > 0), the expression on the left-hand side
of A) becomes
l+b+b*+ .. . +bn
t\b) =

150 ELEMENTARY INEQUALITIES
The preceding proof shows that
f(b) ^ (n+l)l(n—l) for n > 1 and Ъ ^ 1.
Thus, A) is now established for all real a (> 0).
11.11 Show that if a > Ъ > 0, then A < B, where:
A = 1+a+ ■ ■ ■ +аП~1 в = г+5+ ■ • • +bn~1
~ l+a+ . . . +an ' l+b+ . . . +bn
Proof.
1 an 1 bn
1 +
Л l+a+ . . . Ц-а"-1' 5
i.e.,
5~~ + 1/6"+ \jbn~x+ . . . + l/b'
whence it follows that
\\A > 1/5, and so that A < B.
11.12 Prove that
/ 1\" / 1 \re+1
1-| 1 < 1-| (n a natural number).
\ n/ \ n+1/
Proof. We apply the inequality
a1+a2+...+an+an+1 n+y
n+1
with -'
аг = a% ==... = an = l + 1/и, an+1 = 1
to obtain
i
1+
nj и+1

MISCELLANEOUS INEQUALITIES 151
i.e.,
/ 1\" / 1 \re+1
11-| 1 <|1-| 1 (n a natural number).
\ n/ \ n + 1/
Remark by D. С. В. Marsh. This result also follows from 2.30 by
setting b = n, a = w-f-1.
11.13 Let {ak} (k = 1, 2, . . . , 2m) be a set of positive numbers
and {%'} (k = 1, 2, . . ., 2m) be this same set of numbers arranged
in order of magnitude so that
a/ ^ a2' ^ a3' ^ . . . ^ a^m.
Prove L. A. Le Cointe's inequality
a/a2'.. . am'+am+1'am+2' ■ ■ ■ я2т' ^ агаг. . . am+am+1am+2... a2m.
Remark. Concerning this generalized inequality see: T. Popoviciu,
Mathematica, vol. 23, 1947 — 1948, p. 127—128.
11.14 Referred to a rectangular Cartesian coordinate system,
consider the two points with coordinates:
{\{v — u)— W3(y—x), %л/3(у—и)+±{у~х)}, {w—u, z—x},
where x, y, z, u, v, w are arbitrary real numbers. Calculate the
square of the distance between these points and hence derive
the inequality
(x2+y2+z2) — (yz+zx+xy) + (u2+v2+w2)
xl
— (vw-\-wu-\-uv) S; \/3
v y\
w z 1
What conditions must x, y, z, u, v, w satisfy in order for
equality to hold?
Remark. See: H. Langman, The American Mathematical Monthly,
vol. 35, 1928, p. 207.
For two proofs of this inequality (loc. cit.), see: F. Ayres, vol. 36, 1919,
p. 238; and D. R. Curtiss, vol. 36, 1929, p. 289.
11.15 What conditions must be satisfied by the coefficients
A, B, C, D, E, F for the function
Ax2+2Bxy+Cy2+2Dx+2Ey+F A)
to be positive for all real values of x and у ?

152
ELEMENTARY INEQUALITIES
Answer. The function A) is positive for all x and у if and only
if either
or
or
or
1°:
2°:
3°:
A
A
A
> o,
> o,
= в
AB
ВС
AB
ВС
= D
>o,
= 0,
= o,
A
В
D
A
D
С
В,
С
E
В
E
>
D
E
F
0,
>
o,
С
E
o,
A
D
E
F
D
F
>
4°:
= B = C = D = E = 0, F > 0.
Remark. The binomial ax+b is positive for all x if a = 0 and b > 0.
The trinomial ахг-{-2Ьх-\-с is positive for all x if either
a > 0, ac — 62 > 0,
or
a = b = 0, с > 0.
Generalization. What conditions must be satisfied by the
coefficients of the function
A1x*+A2y*+A3z*+2B1yz+2B2zx
+ 2ClX+2C2y+2C3z+D
if it is to assume only positive values for all values of the variables
x, y, zl
11.16 Let f(x) = ao+a1x+ ... +anxn(at ^ 0;i = 0, 1,2,..., и),
g(a;) = /2(s) - &0+6ia;+ . . . +b2nx*n.
Prove Moser's inequality:
b2r+i ^ i/2(l)-
Proof. From the identity
2n in \ 2
2»/= 2mf , (!)
by comparing the coefficients of x2r+1, we find that

MISCELLANEOUS INEQUALITIES 153
2r+l r
K+i = 2 «..«2Г+1-- = 2 2 «и«2г+1-У («* = 0 for г > и). B)
Note the relation
/A) = J «„ = 2 (a,+a2r+1_,) +Sr, C)
!>=0 l>=0
where
On the basis of C) and B), we have in succession
v=0 v=0
r
4 2 я
l-=0
(Solution by D. Djokovic.)
11.17 Prove the inequality
where {afc} is the Fibonacci sequence defined by the relations
a0 — 0, ax = 1, afc+2 = й^+й^+х (A an integer 2g 0).
Remark. Cf. P. S. Modenov: Sbornik zadach po spetsial'nomu kursu
elementarnoi matematiki, Moscow 1957, p. 34, Problem 31.
11.18 If k is a natural number, prove that
=. 1 k+2
Proof. Since
. . . (k+r) > (k+2y-i (r ^ 2),

154 ELEMENTARY INEQUALITIES
then
k+2
which is the required result.
11.19 Prove that
l-a
oo i nl
У — < (n a natural number; a > 1).
11.20 Prove that any real root xx of the equation
xs+px+q =0 (^>, q real numbers) A)
satisfies the condition
^J-4ж1 г ^ 0. B)
Proof. Let x1 be a real root of equation A). Then xx is a
root of the equation
x1 x2+px+q = 0. C)
Since equation C) has real roots, its discriminant must be
non-negative, i.e., the condition B) must hold.
Generalization. Instead of equation A), consider a more
general equation, for example,
zm-\-fiznj\-q = 0 (m, n natural numbers; -p, q real numbers).
(Problem and proof by S. Presic.)
11.21 For what values of x is the following inequality valid:
Vx+1 + V~2x~5?
Answer. The functions лЛс+б, Vx+1, V%x~5 are all real
if x ^ 5/2.
The sense of the inequality remains unchanged if both sides
are squared; thus,
x + 6 > 3x—4:+2V2x2—3x—5 => V2x2—3x—5 < 5—x.
The preceding inequality is meaningful when 5—x > 0; i.e.,
x < 5.

MISCELLANEOUS INEQUALITIES 155
After squaring, the preceding inequality becomes
(ж — 3)(ж+10) < О,
which requires x to satisfy, in addition to the earlier condition
x e[5/2, 5), the condition же(—10, 3).
Conclusion: The relation A) is valid for x e[5/2, 3).
11.22 Solve the inequality
Viz—13 — V3x~ 19 > V5z — 27.
Answer. 19/3 52 x < 9.
11.23 Find the set of points (z, y) in the plane with coordinates
which satisfy the simultaneous system of inequalities:
2x + 3y— 6 < 0, y+2 > 0, ж+l > 0, x2+y2 — 1 > 0.
11.24 Show that if л ^ 0, 6 ^ 0, с ^ 0, rf ^ 0andifc+^ 52 min
(a, b), then
ad-\-bc 52 a&, ac+^ ^ #&•
Proof 1. If
e^ft, A)
the inequality
c+d 52 min(a, &) B)
becomes
c+^ g a. C)
After multiplying the inequalities A) and C) by d and b,
respectively, we obtain
(ad sS bd, bc-\-bd 52 аб) => лй-|-&с 5S a&.
If
igd, D)
the relation B) becomes
c+d<b. E)
From D) and E) we obtain
(be 5S ac, ac-\-ad 52 a6) => ad-\-bc 52 «&.

156 ELEMENTARY INEQUALITIES
In an analogous manner, we may prove that ac-\-bd 5S ab.
Proof 2. Due to Underwood Dudley (The American Mathe-
Mathematical Monthly, vol. 65, 1958, p. 447):
ad-\-bc <S (c+i)max(a, b)
^ min(a, b) max (a, b)
= ab.
Remark. Can this result be generalized?
11.25 Prove the equivalence
\b—c\ < a < b+co [a—c| < b < a-\-c
(a, b, с real numbers).
11.26 Prove that
e
Proof. A) may be written in the form
2x
e<
l\x 2ж+1
ж/ 2ж+2 v '
1°: Let us suppose first that x < — 1 and set a; = —y(y > 1).
Then B) becomes
1
1
1 \ у 1
1 y 1
2«/ У
C)
If Л < В < С (Л, В, С > 0), then log A < log В < log С.
Applying this result to C), we obtain the equivalent inequality

MISCELLANEOUS INEQUALITIES 157
Since
5 a
log(l-e) = - 2 -r (И < 1).
fc=l **
D) takes the form
00 1 / 1 \* °° 1 /1 \* °° 1 Г/1 \* /1
ь=о ^ \2v/ i^^k+l \y/ fc=a ^ L\2// \2i
2yf
E)
The coefficients of (l/2/)fc in the respective sums are
1 1
k-2k> k+1' k\ 2V '
We shall prove that
Consider first the relation
1 1
ft. 2* k+1'
Since
G)
2 > 1+- (A > 1),
the inequality G) is valid.
Let us now examine whether
For k = 2, (8) is true. Let us suppose that it holds for some
&B> 2), i.e., that 2h > k+1. After multiplying by 2, we obtain
2fc+! > 2& + 2 > £+2 (A ^ 2).
The proof of (8) for k ^ 2 may be completed by induction.
Hence, the proof of the validity of F) implies that of B) and
thence of A) for all x < — 1.
2°: Since
— 1— z < — 1 for z > 0,

158 ELEMENTARY INEQUALITIES
B) is true, if we set x = —l—z(z > 0). Thus
22+2 / 1 X-2 2z+l
< 1-ГТТ <e
2z
After multiplication by z/(z-\-l) (> 0), we obtain
2z
e<
1 l\s
1+—
This proves that A) is also valid for x > 0.
11.27 Prove the inequality
2 \ogx~\ogy
Solution. The inequahty
log x—log у 2
x—y x+y
is equivalent to A). Putting x = y+z (z > 0) B) becomes
<» >*•>■»■
B)
Writing ^ instead of г/у (> О) we obtain
The function
is monotonically increasing since
14 t2
f'(t)= = >0 (t>— 1).
' W <+l (^+2J (*+l)(^+2)a V ;
Therefore, we have
/@>/(o) = o (t>o)
which is identical to D).
Remark. This solution is due to D. Djokovic. Another solution can
be found in the paper of B. Ostle and H. L. Terwilliger, Proceedings
of the Montana Academy of Sciences, vol. 17, p. 69—70, 1957.

MISCELLANEOUS INEQUALITIES
159
11.28 Sab ^ -Va2+-Vb2+^(as+b3) (a,b^O).
11.29 min{(b-cJ, {c—aJ, {a-bJ} ^ i(a2
xn—an
11.30
x—a
11.31 sinhr+1 ra+coshr+1 rx < coshr(r+l)a; (x > 0, r > 1).
11.32 smhr+1rx+coshr+1rx>coshr(r+l)x (x > 0, 0< r < 1).
11.33 v %#„ 5= y/a1a2 . . . an ^L |-(й1-|-яге)
{«fc = «1-|-(A — l)d] a1, d ^ 0}.
H-34 n(--l)
11.35
11.36 -
11.37 f^-I
2 я
11.38 J
arctan
л/4-а2
@ < a < 2).
11.39 J
00
11.40 J
ь 2w
2
+ -,,-;-; («> 2)-
11.41
1 5
^ < T