Inequalities (Little Mathematics Library)

Little Mathematics Library

00 P. P. KO ROVKIN

INEQUALITIES

Mir Publishers. Moscow

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LITTLE MATHEMATICS LIBRARY

P. P. Korovkin

INEQUALITIES Translated from the Russian by

Sergei Vrubel

MI.R PUBLISHERS MOSCOW

First Published 1975

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Q English translation, Mir Publishers, t975

CONTENTS

Preface

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6

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12

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19

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23

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Chapter 1. Inequalities

7

1.1. The Whole Part of a Number . . . . . . . . 1.2. The Arithmetic Mean and the Geometric Mean 1.3. The Number . . . . . . . . . . . . . . 1.4. The Bernoulli Inequality . . . . . . . . 1.5. The Mean Power of Numbers . . . . . . . .

Chapter 2. Uses of Inequalities

32

2.1. The Greatest and the Least Function Values 2.2. The Holder Inequality . . . . . . . . . . .

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32

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40

2.3. The Use of Inequalities for Calculation of Limits . . . . 2.4. The Use of Inequalities for Approximate Calculation of Quantities . . . . . . . . . . . . . . . . . . . .

43

Solutions to Exercises

58

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49

5

PREFACE

In the mathematics course of secondary schools students get acquainted with the properties of inequalities and methods of their solution in elementary cases (inequalities of the first and the second degree).

In this booklet the author did not pursue the aim of

presenting the basic properties of inequalities and made an attempt only to familiarize students of senior classes with some particularly remarkable inequalities playing an important role in various sections of higher mathematics and with their use for finding the greatest and the least values of quantities and for calculating some limits. The book contains 63 problems, 35 of which are provided

with detailed solutions, composing thus its main subject, and 28 others are given in Sections 1.1 and 2.1, 2.3, 2.4 as exercises for individual training. At the end of the book the reader will find the solutions to the given exercises.

The solution of some difficult problems carried out individually will undoubtedly do the reader more good than

the solution of a large number of simple ones.

For this reason we strongly recommend the readers to

perform their own solutions before referring to the solutions

given by the author at the end of the book. However, one should not be disappointed if the obtained results differ from those of the patterns. The author considers it as a positive factor.

When proving the inequalities and solving the given

problems, the author has used only the properties of inequa-

lities and limits actually covered by the curriculum on mathematics in the secondary school.

P. Korovkin. 6

CHAPTER 1

Inequalities

The important role of inequalities is determined by their application in different fields of natural science and engineering. The point is that the values of quantities defined from various practical problems (e.g. the distance to the Moon, its speed of rotation, etc.) may be found not exactly, but only approximately. If x is the found value of a quanti-

ty, and Ox is an error of its measurement, then the real value y satisfies the inequalities

x - I AxI x. Thus, [x] is the integer (whole number) defined by the inequalities

[x] < x C' [x] +,I .. 7

For example, from the inequalities

3 n. Proof. Use the method of mathematical induction'. First of all check up the validity of the theorem for n = 2, i.e. show that

x1x2=1=x1+x2>2. Solving the question, examine the two given cases separately: x1 = x2 = 1.

(1)

In this case xl + x2 = 2, and the theorem is proved.

0 < x1 < x2 Here xl < 1, and x2 > 1, since their product is equal to 1. (2)

From the equation

(1 - x1) (x2 - 1) = x2 + x1 - x1x2 - 1 it follows that (4) x1 + x2 = x1x2 + 1 + (1 - x1) (x2 The equation (4) has been established without limitations to

the numbers x1 and x2. Yet, taking into account, that x1x2 = 1, we get

x1-+2=2+(1-x1)(x2-1). At length, since x1 < 1 < x2, then the last number is positive and x1 + x2 > 2. Thus, for n = 2 the theorem is already proved. Notice, that the equation

x1+x2=2 is realized only when xl = x2. But if x1

x2f then

x1+x2>2. Now, making use of the method of mathematical induction, assume that the theorem is true for n = k, that is, sup1 More detailed information concerning mathematical induction

is published in the book by I. S. Sominsky "The Method of Mathemati

cal Induction", Nauka, Moscow, 1974.

13

pose the inequality

x1+x2+x3+. .. +xh>k . . . xh = 1, and prove the theorem for n = k + 1, i.e. prove that

occurs, if x1x2x3

X1 +X +x3+ . . . +xh+xh+1>k+ 1, if x1x2x3

.

.

1, for x1 > 0, X1>01 x3 > 0, ... xh>0, xk+1>0.

. xkxk+l =

First of all, it is necessary to notice that if . xkxh+l = 1, then there may be two cases: (1) when all the multipliers x1, x2i x3, ..., xh, xk+1 are X1X2X3 .

.

equal, that is

x1 = x2 = x3

` xh = xk+1,

(2) when not all multipliers are equal. In the first case every multiplier is equal to unity, and their sum equals k + 1, that is

x1+x2 +x3+ . .. +xh+xh+1 =k+

1.

In the second case, among the multipliers of the product . xkxh+l, there may be both numbers greater than

X1X2 .

.

unity and numbers less than unity (if all the multipliers were less than unity, then their product as well would be less than unity). For example, suppose x1 < 1, and xk+l > 1. We have (xlxh+l) X2x3 .

.

. xh =

1.

Assuming yl = xlxk+lf we get y1x2x3 . . . xh = I. Since here the product k of positive numbers is equal to unity, then (according to the assumption) their sum is not

less than k, that is

y1+x2+x3+. .. +xh>k.

But

x1+x2+ x3+

+xh+xk+1 =

= (yl+x2+x3+. . +xk)+xk+1-y1+xl>

k+xk+l-yl+x1=(k+1)+xh+l-yl+xl-1

14

Remembering that y1 = xlxk+l we get XI + X2 + X3 + . . . + xk + xk+1 >

>(k+ 1) +xh+l - xlxk+l+xl - 1 =

=(k+1)+(xk+1-1) (1-x1).

Since xl < 1, and xh+l > 1, then (xh+l

and, hence,

xl+x2+x3+

- 1) (1 - x1) > 0

+xk+xk+1>

>(k+1)+(xh+l-1)(1-xl)>k+1.

Thus the theorem is proved. Problem 1. Prove, that if x1, x2, x3i ..., x, are positive numbers then x2 } S3

}

... + xn'

m x!

n,

the equality being valid only when

x1=x2=x3= ... =xn. Solution. Since x1 X2

x2 x3

...

xn-1 xn

xn x1

then the inequality follows from Theorem 1, the sign of equality holds only when xn

x1

namely, when x1 = x2 = x3 = . . . = xn. Problem 2. Prove the inequality x2+2

Solution. We have X2

2

_ x21

V52+1 1

2.

_Yx2+1+

1

j/x2+1 + /x2+1 /x2+1 Since the product of addends in the right-hand member of the -/x2+_j

equality equals unity, then their sum is not less than 2.

The sign of equality holds only for x = 0.

Problem 3. Prove that for a > 1 log a + logo 10 > 2. 15

Solution. Since log,, 10 -log a = 1, then log a + toga 10 - log a -f - toga > 2.

Problem 4. Prove the inequality z2 1-{-x4

1

2

Solution. Divide by x2 the numerator and denominator of the left-hand member of the inequality: X2

1+x4

=

1

a 1

+xa

x

Since

x

-- x2> 2 and, hence,

x2= 1, then

X

1

1

z2+

X2

2

Now let us prove the statement made at the beginning of the section. Theorem 2. The geometric mean of positive numbers is not greater than the arithmetic mean of the same numbers. If the numbers x1, x3i . ., xn are not all equal, then .

the geometric mean of these numbers is less than their arithmetic mean. Proof. From the equality g = v xix2 ... X" it follows

that

2n

x1 x2 9

9

or x1 x2 9

9

9

xn 9

=1.

Since the product n of the positive numbers equals 1,

then (Theorem 1) their sum is not less than n, that is 9

9

9

Multiplying both members of the last inequality by g and

dividing by n, we get a= xi+x2+...+xn n 16

g

Notice, that the equality holds only when ?r = x2 = . 9

9

=

= 1, that is x1 = x2 = ... == xn = g. But if the numbers g x1, x2i ... , xn are not equal, then

avg.

Problem 5. From all parallelepipeds with the given sum

of the three mutually perpendicular edges, find the parallele-

piped having the greatest volume. Solution. Suppose m a + b + c is the sum of the

edges and V = abc is the volume of the parallelepiped. Since

V=3 abc< a +b3 } 3

then V- (

g

)

n

a

= ca.

So the first part of the inequality (6) is proved; the second is proved in a similar way. p" From the inequality (6) it follows, in particular, that the harmonic mean c_1 does not exceed the arithmetic mean c1.

Problem 8. Prove that if a1, a2, ..., an are positive

numbers, then

(a1 + a2 + ... +an)

( a1

2 Solution. Since c_1 0, ..., a,, > 0.

(7)

Solution. Since the geometric mean does not exceed the arithmetic mean, then n ... ann n+ n a1a2 ... an = V V na1a2 Multiplying both members of this inequp1}ty by n, we shall

get the inequality (7). From the inequality (7) it follows, that

2a1a2 < a; +a2, 3aja2a3 G a'+ a,+ a3, 4aja2a3a4 < a; + az + a3 + a4,

that is, the doubled product of two positive numbers does not exceed the sum of their squares, the trebled product of three numbers does not exceed the sum of their cubes and so on.

1.3. The Number e The number e plays an important role in mathematics. We shall come to its determination after carrying out the solution of a number of problems in which only Theorem 2 is used. (a

Problem 1. Prove that for any positive numbers a, b, b) the inequality

n+1/abn a+nb n+1

is true. Solution. We have

n

n+/abn = n+V abb ... b < a - b b ... n+1

a { nb

}b

n+1

n

and that suits the requirement. Problem 2. Prove that with the increase of the number n the quantities

_(

1

xn-(1+n)n

and

1

zn1n)n

2*

19

increase, i.e. n+1

1

xn < xn+1 = (1 + n+1

)

, 1

\\n+1

znyn+1= (t + n+1

)n+2

Solution. We have

yn=(t+

1

n

)n+1

n -i-1 )n+1 n

=

n

n+1

( n+1)

_

1

(

(se e

1

)fl+1

1

Zn+1

n+1 designations of Problem 2). Since zn increases with

the increase of the number n, then yn decreases.

In Problems 2 and 3 we have proved that x1= (t + 1 )1 =2<x2= (t + 2 )2=

=2.25<x3< ... <xn< ..., y1= (t + 2 )2==4>y2=

(t + 1)'=3.375>Y3> 20

.. > yn > ...

On the other hand,

2 =x1<xn= (1 + n )n
0, this value being equal to -

4

In case of arbitrary value of a > 1 the theorem is proved by using the inequality (12), demonstrated in Theorem 3.

Since a > 1, then

(1+z)a+az, z

the equality holding only when z = 0. Assuming here, that 1 + z = y, we get

ya>1 +a(y-1), ya - ay>I -a, y>0,

the sign of equality holds only when y = 1. Multiplying both members of the latter inequality by ca, we get (cy)a - aca'1 (cy) > (1 - a) ca, y > 0. Assuming

x=cy and we get

aca-1=a,

-) a

c= ( a

1

as-1 a

a

xa-ax>(1-a)ca=(1-a) (a ) a-1

A

here the equality occurs only when x = c = (a) a Thus, the function

xa - ax,

a=1

a >1, a>0, x>0, 1

takes the least value in the point x=

a

equal to

a

(1- a) ( a) 3-0888

a-1

. The theorem is proved. 33

In particular, the function x2 - dx (a= 2) takes the 1

least value in the point x = (2 )L z

(1-2) ( a ) 2-1

=-4

.

1

= a

equal

,

to

This result is in accordance with

the conclusion, obtained earlier by a different method. The

function X3-27x takes the least value in the point 3

1

x- (2) 3-1 =3, equal to (1-3) (2)

3-1

=

54.

Note. Let us mark for the following, that the function

ax -,xa = -(xa - ax), where a > 1, a > 0, x > 0, takes the greatest value in the point X= 1

I

a

equal to

(a-1)(a l

Fig. 1

a

Problem 1. It is required to saw out a beam of the greatest durability from a round log (the durability of the beam is directly proportional to the product of the width of the beam by the square of its height). Solution. Suppose AB = x is the width of the beam, BC = y is its height and AC = d is the diameter of the log (Fig. 1). Denoting the durability of the beam by P, we get

P = kxy2 = kx (d2 - x2) = k (d2x - x3). The function d2x - x3 takes the greatest value when 1

` X= (d2 3-1 3

J

--

d

y2=d2-x2= 23 d2,

y= d1/9 Y2=xV2 34

Thus, the beam may have the highest (greatest) durability if the ratio of its height to its width will be equal to V/2 x

1.4-

5.

Problem 2. Find the greatest value of the function y = sin x sin 2x. Solution. Since sin 2x = 2 sin x cos x, then sin x sin 2x= = 2 cos x sine x = 2 cos x (1 - cost x) = 2 (z - z3), where z =:cos x and, hence, -1 < z < 1. The 'function z - z3 =

Z (1 - z2) takes a negative value when -1 < z < 0,

Fig. 2

is equal to 0 when z = 0 and takes a -positive value when 0 < z < 1. Therefore, the greatest value of the function is gained in the interval 0 < z < 1. It is shown in Theorem 5 that the function z - z3, z > 0, takes the greatest value in the point

Z=(13) 3-1

-

'1

In this point

sin x sin 2x = 2z (1- zz) =

V-3 ( 1

- T3) = 3 i/3

So, the function y sin x sin 2x takes the greatest value in those points, where z cos x = _ and this value is 3

equal to 0.77. The graph of the function y = 3 _/3 = sin x sin 2x is shown in Fig. 2. Problem- 3. Find the greatest value of the function y = cos x cos 2x. 3*

35

Solution. The function y = cos x cos 2x does not exceed

1, since each of the cofactors cos x and cos 2x does not exceed 1. But in the points x = 0, ±2n, ±4n, .. . cos x cos 2x = 1.

Thus, the function y = cos x cos 2x takes the greatest value of I in the points x = 0, ±2n, ±4n, . The graph of the function y = cos x cos 2x is drawn in Fig. 3.

...

Fig. 3

Problem 4. Find the least value of the function

xa + ax, where a ] 0, a < 0, x >,O. Solution. Since a < 0, then according to the inequality (12)

(1 + z)a > 1 + az, and the sign of equality holds only when z = 0. Assuming

1 -{-z=y, z=y-1, we get

ya>1+a(y-1), Y>0, the sign of equality occurring only when y = 1. From the last

inequality it follows, that

y17 -ay>1 - a, (cy)"-aca_'(cy)>(1

a) ca.

Assuming a = -aca-1, x = cy, we get a

A a

) a-1 1

the equality holding only when x = c 36

(

a

) a= 1

Thus, the function xa -{- ax takes the least value in the point

X='

as )a 1'

a

a

equal to (1 -a)

) a-1 .

For example, the function

+27x, x> 0, takes the least value in the point 1 1

27

3

-1

1

3

This value equals 3

(14-

31 / 1

27

1 _1

- 3

=4.

3

Problem 5. Find the optimum dimensions of a cylindrical tin having a bottom and a cover (dimensions of a vessel are considered to be the most profitable, if for a given

volume the least amount of material is required for its manufacture, that is, the vessel has the least surface area).

Solution. Let V = nr2h be the volume of the vessel, where r is the radius, h is the height of the cylinder. The total surface area of the cylinder is S = 2nr2 + 2nrh. Since h = V

, then

S = 2nr2 + 2nr r2 = 2nr2 + 2V r 1 Assuming x = T , we get

.

$=2nx 2+2Vr=2n (X-2 a7

The function x 2+ V x, according to the solution of the previous problem, takes the least value when

X= / V

-2-1

l2n

;

/

V V

Returning back to our previous designations, we find 1

_

2n

,r3=

713

V

nr2h

h

2a= 2a 'r-2

h = 2r = d.

Thus, the vessel has the most profitable dimensions, if the height and diameter of the vessel are equal.

Exercises

6. Find the greatest value of the function x (6 - x)2

when 0<x 0, the following inequality

is true

3x 3, the following inequality is true

rn>n+/n +1. Indication. Make use of the inequality (8). 13. Find the greatest of the numbers

1, -1/12, V3, V4, y5, ..., fn, ...

.

14. Prove the inequality

rn1+al+a2-}-...+an,

if the numbers ai are of the same sign and are not less

than -1.

16. Prove the inequality (aib1 + a2b2 + ... + anbn)2

(a,+a2+...+a')(b;+bz+... +bn).

(19)

Indication. First prove, that the polynomial (a1x - b1)2 -I- (a2x - b2)2 +

. .

x2(a2+a2+

. + (a,,x -

b,,)'

+an) -

- 2x (a1b1 + a2b2 + ... + a,, b,,) + + (b i + b z -{-

. .

. + bn )

cannot have two different real roots. 17. Using the inequality (19), prove, that the arithmetic mean is not greater than the root-mean-square. 39

18. Prove the inequality

_n 1, 1 + 4 =1, x>0, y > 0, then

xy < P -{ 4

(20)

Proof. By virtue of Theorem 5, if a < 1, a > 0, x > 0, then

xa-ax> (1-a)

a

a a-1

a

Assuming in this inequality that a = p, a = py, we get P

P

XP-(Py)x>(1-P) (py )' 1=(1-P)yP-'. Since P -}- 4 =1, then q 40

=1-

P`p-

p

, q= ppq

,

P-1=

(21)

Putting these values into the inequality (21), we get

xn-pyx> - 9 y4. Dividing all the members of the latter inequality by p and transposing the negative members to the opposite side, we get the inequality (20). Theorem 7. If a1, a2f ..., an, b1, b2i ..., bn are positive numbers, and p and q satisfy the conditions of Theorem 6, then

ajb1 + a2b2 + ... + anbn 1

(a' + a2 -I-

1

... + an) p (b? + b2 + ... + bn) q

(22)

Proof. Suppose

ai+a2+ ... +an=Ap, bi I bz+ ... + bn=B4. Then the right member of the inequality (22) will be equal to 1

1

(Ar) P (B4) 4 = AB.

Now suppose a1 = Act, a2 = Act, ... , an = Acn, b1= Bd1i b2 Bd2, ... , bn -{- Bdn. Since

AP=api +a2+ ... + an =APcPj +APcP2 +...+Apcn=Ap(ci+cP2 +...4cn), then

In a

Ci-f-CZ+...+cn=1. similar way, it is checked that

dI+d2+...+dn=1. Now using the inequality (20), we get

a1b1= AB (c1d1) x3 >

. . .

> xn, then 0 0. On the other hand, since

A =x1- (x2-x3+xa-.. . + (-1)n 2xn)

and the quantity in brackets is positive too, then A < x1. Thus, the lemma is proved. Lemma 2. If 0 < a < 1, then the following inequalities are true (2n+1)1_0'-(n+1)1-"

1-a

< (n+1)a + (n+2)a + ... 1

... +

1

(2n)0'


1. By virtue of Theorem 5, the greatest value of the function

a y-ya is 1

1

1

a

(a-1) a

1

a

1aa

(a

Multiplying the last quantity by a, we shall find the greatest value of the function a

(1-a)

aaa )all

1

1

a y - ya which is, hence, equal to a

a

+ a

1 1

a

= ( 1-a ) ( aa)

a-1

61

It. The function i/x - 2x, x> 0, a = 4

,

a = 2, bas the

greatest value, equal to 4

4-1

4/

3

1

3

1

`4f I

,0 the following inequality is true Therefore, for all x>

x - 2x < 8 , or Vx < 8 -I- 2x. 12. Write down the inequality (8) in the form of

(nn1)n<e, If n > 3 > e, then 0. Raising to a power of n we get n)n 12 r +a n=(1+a)n

.(12

Assuming that n

2, 2

taking Theorem 3 as the

basis, we get n

n>(1-{ 2 an)=1+nan+42an. 2

62

Hence, it follows that

n> n2 an, an
-1, the inequality is obvious

1+a1>1+ a1. Let us assume, that the inequality is true for n = k, that is (1 + a1) (1 + a2)

. . .

(1 + ak)

>1+a1+a2+...+ak.

Multiplying both members of the inequality by (1 + ak+1), we get (1 + a1) (1 + a2) ... (1 + ak) (1 + ak+1)

>(1+a1+a2+.. +ak)(1+ak+1)

= 1 +a1+... +ak+ak+1 +alak+1+ + a2ah+l +

.

.

.+

akak+-1

Since the numbers a1i a2, ..., ak, ak+l are of the same sign, then alak+l + a2ah+l + . + aka k+ l > 0 .

and, therefore, (1 + a1) (1 + a2) ... (1 + a,,) (1 + ak+1)

>1 + a1+ a2 + ... + ak +ak+1,

that is, the inequality is proved also for n = k + 1. 63

This finally proves the inequality to be true

(1 + ai) (1 + a2) ... (1 + an)

+a1+a2+... ma,,

for all n. 16. If the polynomial (a1x - b1)2 + (a2x - b2)2 + .. . . + (an,x - bn)2 has a true root x - x1, that is

. + (anx1 - bn)2 = 0, then every number a1x1 - b1, a2x1 - b2, ..., anx1 bn is equal to zero, that is, 0 = a .1x1 - b1 = a2x1 - b2 = ... = a,,xl - bn, (a1x1 - b1)2 + (a2x1 - b2)2 +

b1

x1 =

a1

=

.

.

b,, a,,

b2 a2

Thus we proved that the polynomial b1)2 + (a2x - b2)2 + ... + (anx - bn)` (alx x2 (a2 + a2 + . . + an) -

-

.

2x (alb1 + a2b2 +

... +

anbn) +

+ (b2 + b2 T . . + bn) cannot have two different true roots and, therefore, (albs + a2b2 + ... + anbn)2 .

- (a, + + a 'n) (b i + . From this follows the inequality (19) (albs + a2b2 +

. .

-4-- b

n
0, then a + 1 > 1 and, hence, 1

(1+ n )1+a>1+1+a n 67

Multiplying these inequalities by nl+a, we get

(n+1)'+a> nl+a+(1 +a)na, (n-1)1+a>n1+a-(l +a)na

From these inequalities it follows that nl+a-(n-1)1+a

< na