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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CONTENTS
Preface
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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
When solving practical problems, it is necessary to take into account all the errors of the measurements. Moreover, in accordance with the technical progress and the degree of complexity of the problem, it becomes necessary to improve the technique of measurement of quantities. Considerable errors of measurement become inadmissible in solving
complicated engineering problems (i.e., landing the mooncar in a specified region of the Moon, landing spaceships on the Venus and so on).
1.1. The Whole Part of a Number The whole (or integral) part of the number x (denoted by
[ x ]) is understood to be the greatest integer not exceed-
ing x. It follows from this definition that [x] < x, since the integral part does not exceed x. On the other hand, since [x] is the greatest integer, satisfying the latter inequality, then [x] + 1 > 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] - 3, 1 3
2, [5] = 5.
5,
The ability to find the integral part of a quantity is an important factor in approximate calculations. If we have the skill to find an integral part of a quantity x, then taking [x] or [x] + I for an approximate value of the quantity x, we shall make an error whose quantity is not greater than 1, since
0<x-[x]<[x]+1-[x]=1, 0<[x]+1-x<[x]+1-[x]=1.
Furthermore, the knowledge of the integral part of a quanti-
ty permits to find its value with an accuracy up to 12 The quantity [x] + 2 may be taken for this value. Yet, it is important to note, that the ability to find the whole part of a number will permit to define this number and, with any degree of accuracy. Indeed, since [Nx]
[Nx]
Gx
[Nx] + 1 N
N
Thus, the number [Nx]
1
-)V + -IN1
With large N the error will be small. The integral part of a number is found in the following problems. differs from the number x not more than by
N, .
Problem 1. Find the integral part of the number
8
Solution. Let us use the following inequalities
1<1 <1,
0.7<j/Z <0.8, 0.5 < V < 0.6,
0.5 < 4 <0.5,
0.4
<1+0.8+0.6 d-0.5+0.5, that is, 3.1 < x < 3.4, hence, [x] = 3.
In this relation, it is necessary to note that the number
3.25 differs from x not more than by 0.15. Problem 2. Find the integral part of the number 1
1
1
1
x1000000
Solution. This problem differs from the previous one
only by the number of addends (in the first, there were only 5 addends, while in the second, 1000, 000 addends). This
circumstance makes it practically impossible to get the solution by the former method. To solve this problem, let us investigate the sum
and prove that
21/n- 1-2Yn< Indeed, since
Vn
<2yn -2Vn-1.
2
Vn a
and
YnT-1>Yn, it follows that
21/n+1-21/n <
-
2
1
Thereby proof has been made for the first part of the inequality (1); its second part is proved in a similar way.
Assuming in the inequalities (1) n = 2, 3, 4,
we get
21/3-21/2 <
..., n,
<2y-f-2,
1
<21/3-2Y2,
21/4-21/3 <
21/5-21/4<
-VT 1
<21/4-21/3,
...........................
2V
1-21/n<
1
<21/n-21/n-1.
Vn Adding these inequalities, we get
21/n-{-1-21/2 < < 1/12
... } Vn <2 j/n-2.
v3 }
}
Adding 1 to all parts of the obtained inequalities, we find
21/n-{-1-2V2+1< < 1-}-
1 + v3 1+
1
... T
1
vn
<2y-n-1.
(2)
Since 21/2 < 3, and 1/n -{-1 > 1/n, it follows from the inequalities (2) that
21/n-2<1+
v2
+ r/3 +
1
+... + vn < 2 1/n =1.
*9
(3)
Using the inequalities (3) we can easily find the integral part of the number Y
2
1
1
1
+ 1000 000 -/2 + V-3 + JA + Thus, taking in the inequalities (3) n =1000 000, we get 2 }V1000 000 - 2 <
=+
<1{
-} V3 -f
+
1000 000
1000 000 -1,
<2
or
1998 < y < 1999.
Hence, [y] = 1998.
From the inequalities (2) it follows that the number 1998.6 differs from y not more than by 0.4. Thus, we have 40 calculated the number y with an accuracy up to 7_998.4 % = 0.02 %. The numbers 1998 and 1999 differ from the number y not more than by unity, and the number 1998.5 differs not more than by 0.5. Now let us examine the next problem of somewhat different pattern. Problem 3. Prove the inequality 99 100
3 , 5 6
1
X - -2-, 4 Solution. Suppose 2
4
Y3
1
10
100
6
7 "' 101
Since 2
1
3
4
5
100
99
6
4<5, 6<7100<101
T<,
it follows that x < y and, consequently, x2 <
xy
_
1
2
3
4
5
99
6
100
_
1
2'3'4 6' 7 101 101 Finding the square root of both members of the inequalities yields 100.
x<
<0.1.
101 ii
Exercises 1. Prove the inequalities
+... ...+ Vn- <2Yn-2Ym-1.
2Yn+1-2Ym
1
2. Prove the inequalities
1,800 < 10,000
+ 1/10,001 +
+
x/1,000,000
< 1,800.02.
3. Find [50z], where z
+ 1x/1 0,001 + ... + x/1,000,000 _
]/10000
Answer. [50z] = 90,000.
4. Prove the following inequality using the method of mathematical induction 1
3
2n-1 <
5
2 ' 4' 6 "'
2n
1
3n -1
5. Prove the inequality 1
3
5
99
1
2*4'6 "' 100[12 1.2. The Arithmetic Mean and the Geometric Mean If x1, x2, ..., xn are positive numbers, then the numbers formed with them a = x1 x2'}'n ... +an g = n x1x2 ... Xare called, respectively, the arithmetic mean and the geometric mean of the numbers x1, x2i . . ., xn. At the beginning of the last century, the French mathematician 0. Cauchy has established for these numbers the inequality g < a,
often used in solving problems. Before proving the inequality we have to establish the validity of an auxiliary assertion 12
Theorem 1. If the product n of the positive numbers tj, x21 ..., xn is equal to 1, then the sum of these numbers is not less than n: xlx2f . . ., xn = 1 = x1 + x2 + . . . + xn > 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-<_ 27
.
m 3
c
'
The sign of equality holds only when a
= b-= c=7, that is, when the parallelepiped is a cube. Problem 6. Prove the inequality
n!<(n21)n,
(5)
Solution. Using Theorem 2, we get n o t= 1.2.3 ... n< 1+2+3+... -i- n (n+ 1) n
n+1
2n
2
n
-
Raising to the nth power both parts of the last inequality, we get the inequality (5). Definition. The number
+an)a
Ca- a1+a2 n
is termed the mean power of numbers a1, a2, ..., an of the order a. Particularly, the number
_
a1-- a2+... { an n
Cr
is the arithmetic mean of the numbers ar, a2, ..., an, the number C2
- (a
-{- a22 -F 1
2-0866
n
... -{- an
2
) 17
is named the root-mean-square, and the number a11+a21... { _
ant)-1n
C'1
n
1
1
1
+---+ an Q1 a2 is called the harmonic mean of the numbers a1, a2, ..., an, }
Problem 7. Prove that if a1, a2i ..., an are positive numbers and a < 0 < 3, then (6) ca < g < c o, that is, the mean power with a negative exponent does not exceed the geometric mean, and the mean power with a positive exponent is not less than the geometric mean. Solution. From the fact, that the geometric mean of positive numbers does not exceed the arithmetic mean,
we have
a
a1+a2+...+an
a a ... any n ala2 Raising both parts of the last inequality to a power 1 and
taking into consideration, that
< 0, we get 1
a1+a2+...-f-an
a1a2 ... an > (
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
-{- ... + n) , n2.
< a1+a2+...+an
n
n
--F-H-...+-an a1 a2 1
1
1
=Cj.
it follows from this inequality that n2 < (at + a2 + ... - an) 18
(
a+ a+ ... + an
Problem 9. Prove the inequality
nala2 ... a, < ai -}- a2 + ... H- an, where a1 > 0, a2>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
zn
Solution. Setting in the inequality of the previous prob-
lem a=t, b=t+ n+1
t
1
we get 1+n
1
t
,
1
n
1
Raising both parts of the inequality to the (n + 1)th power,
we shall obtain
(t + 1 ) n < (t + n+ 1) n+1 , that is xn < xn+i 1
1
The second inequality is proved in a similar way.
Problem 3. Prove that
yn=(t+
1 1n+1 n )
decreases with the increase of the number n, that is 1
Yn>yn+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<
n In+t-yn
Thus, the variable xn satisfies two conditions: (1) xn monotonically increases together with the increase of the number n;
(2) xn is a limited quantity, 2 < x, < 4.
It is known, that monotonically increasing and restricted
variable has a limit. Hence, there exists a limit of the variable quantity xn. This limit is marked by the letter e, that is,
e= Jim x= lim(1+ n-.oo
n-.oo
-) 1
n.
n
As the quantity xn increases reaching its limit, then xn is smaller than its limit, that is
xn= (1
+ n )n<e.
(8)
It is not difficult to check that e < 3. Indeed, if the num-
ber n is high, then xn
1
(1 ± 5) = 2.985984.
e = lim xn <, 2.985984 < 3. n-.oo
In mathematics, the number e together with the number at is of great significance. It is used, for instance, as the base of logarithms, known as natural logarithms. The logarithm of the number N at the base e is symbolically denoted by
In N (reads: logarithm natural N). It is common knowledge that the numbers e and at are irrational. Each of them is calculated with an accuracy of up to 808 signs after the decimal point, and e = 2.7182818285490 .
.
.
.
Now, let us show that the limit of the variable yn also
equals e. Indeed,
limy"=lim(1+n)n+1 .1im(1+1)n(1
n e-1 =e. 21
Since y,ti diminishes coming close to the number e (Pro-
blem 2), then
i 1) 1
n+i
>e.
(9)
Problem 4. Prove the inequality
n! > (en y.
(10)
Solution. We shall prove the inequality (10) using the
method of mathematical induction. The inequality is easily checked for n = 1. Actually,
11=1>
e
Assume, that the inequality (10) is true for n = k, that is k!>(e)k.
Multiplying both members of the last inequality by k+ 1, we get (k+1)!>(k)k(k+1)=( keg `k+1
(k+ 1) k! =
e1
+k)
J
k
)k
Since, according to the inequality (8) (1 +
(k+1 )k+1
(k+1)!> (k71 )k+1 e e
< e, then k
e
e
that is the inequality (9) is proved for n = k + 1. Thus the inequality (9) is proved to be true for all values of n. Since e < 3, it follows from the inequality (9) that
n! > ( 3 )n. By means of the last inequality, it is easy to prove that 300! > 100300
Indeed, setting in it n = 300, we get 3001 > 22
( 330) 300
_
100300.
The inequality
n!<e(
n+1 )n+1 e
is proved completely the same way as it is done with the inequality of Problem 4. 1.4. The Bernoulli Inequality In this section, making use of Theorem 2 we shall prove the Bernoulli inequality which is of individual interest and is often used in solving problems.
Theorem 3. If x > -1 and 0 < a < 1, then (1 +x)a + ax. However if a < 0 or a > 1, then
(1+x)01 >1+ax.
(11) (12)
The sign of equality in (11) and (12) holds only when x = 0.
Proof. Suppose that a is a rational number, bearing in mind that 0 < a < 1. Let a = , where m and n are positive integers, 1 < m < n. Sincen according to the condition, 1 + x > 0, then M
(1+x)a=(1+x) n =V(1+x)m,1n-m
_
=V(1+x)(1+x) ...
1
m
n-m n
m (I ±x)+n-m
n+mx
n
n
1
X. 1 + ax. +m n
The sign of equality occurs only when all multipliers stand-
ing under the root sign are identical, i.e., when I + x = 1, x = 0. But if x 0, then
(1 + x)a < 1 + ax. Thus, we have proved the first part of the theorem considering the case, when a is a rational number. 23
Assume now, that a is an irrational number, 0 < a < 1.
Let r1, r2,
.
.
., r,
.
.
.
be the sequence of rational numbers,
having for a limit the number a. Bear in mind that 0 < < r < 1. From the inequalities
(1 + x) rr <1 + r,,x, x > -1, n = 1, 2, 3,
...,
already proved by us for the case when the exponent is a rational number, it follows that
(1-J x)a=- Jim (1+x)'n
rn-.a
Thus the inequality (11) is proved for irrational values of a as well. What we still have to prove is that for irrational
values of a when x* 0 and 0 < a< 1 (1 + x)a < I -r- ax,
i.e., that when x 0 in (11), the sign of equality does not hold. For this reason, take a rational number r such that a < r < 1. Obviously, we have
(1+x)a=[(14_x)r]r. Since 0
a < 1, then as it has already been proved (1 -Fx)r
+
X.
Hence,
('J+x)a<(1 + a x)r. If x
x) r < I +r a x- 1 -;- ax, that is
0, then (1 -F-
(1 +x)a <1 + ax. Thus the first part of the theorem is proved completely.
Now, move on to proving the second. part of the theorem.
If 1 + ax < 0, then the inequality (12) is obvious, since its left part is not negative, and its right part is negative. If 1 -+- ax > 0, ax > -1, then let us consider both cases separately.
Suppose a > 1; then by virtue of the first part of the theorem proved above we have 1
(1+ax)a <1+ a ax=1+x. 24
Here the sign of equality holds only when x = 0. Raising
both parts of the last inequality to the power a we get I -- ax <' (1 + x)a.
Now let us suppose a < 0. If 1 + ax < 0, then the inequality (12) is obvious. But if I + ax > 0, then select the positive integer n, so that the inequality - n < I would be valid. By virtue of the first part of the theorem we get a (1
{
x)
n
<1-a
x,
(1+a1--x >1 {
a n
x
n
2
the latter'inequalit.y is true, since 1 > 1- n2 x2) . Raising both parts of the latter inequality to the nth power we get
+n x)n>1+nn x=1+ax.
(1
Notice, that the equality is possible only when x = 0. Thus, the theorem is proved completely.
Problem 1. Prove, that if 0 > a > -1, then (n+1)a+1_na+1
na+1(n-1)a+1 +1
(13)
Solution. Since 0< a +1 <_ 1, then according to the
inequality (11) we have
(1 +
n
`a+1 < 1+ an 1 (1-1)a+'<1-a+1
n
n
Multiplying these inequalities by na+1 we obtain (n+1)a+1
(n-1)a+1 < na+1 - (a-{-1) na.
The inequalities (13) easily follow from these inequalities. 25
Problem 2. Prove, that if 0 > a > -1, then
(n+1)a+1-ma+1 a+1
-}-na
<ma+(m+1)a+ ... Setting
Solution.
in
m+1, ..., n, we get
(m+1)1+a-m1+a
the m
1+a
(+1)a+i
inequalities (13)n= m, m1+a-(m-1)1+a
1+a (m+1)1+a-mi+a
(m+2)1+0'-(m+1)i+a
1+a
(m+3)1+a-(m+2)1+a
(14)
<(m-1)a< 1+a < (m+2)'< (m+2)1+a-(m+1)1+a
1+a
1+a
................................. (n+1)1+a-n1+a
1+a
Problem 3. Find the integral part of the number + 1 + x 35 3s ... + 31,000,000 1
1
33
Solution. Setting
in the inequality (14) m = 4, n =
=1,000,000, a = - 3 , we get 2
2
2
2
1,000,013 -43 <x< 1,000,002 3 -33
that is 2
2.1,000,0013
- 2.4 3 <x<2. 1,000,000 3 - 2 ,3 3 2
2
2
.
Since 2
2
2
. 1,000,001 3 > 2 . 1,000,000 3 =2.10,000=15,000,
2 VT-6 -
V54 < 4, 2 V9 > 2 V8 = 3,
then
15,000 -4 < x < 15,000 - 3, that is 14,996 < x < 14,997. From these inequalities it follows that [xl - 14,996. 26
1.5. The Mean Power of Numbers In Sec. 1.2 before Problem 7 we have already named the number
i as { a2 + ... -{- an a Ca==
n
)
the mean power of order a of the positive numbers a1i a2, ..., an. In the same problem, it has been proved, that Ca
if a< 0< P.
Here, should be proved the validity of the inequality Ca < co any time when a < P. In other words, the mean
power of order a is monotonically increasing together
with a. Theorem 4. If a1, a2, ..., an are positive numbers and
a<
then ca < cs, and ca = cg, only when a1 = a2 =
= an.
have Proof. For the case, when the numbers a and different signs the theorem has been proved above (refer to Problem 7, Sec. 1.2 and the definition prior to it). Thus, we have to prove the theorem only for the case when a and P have the same signs.
Assume, that 0 < a < 3, and let
1
k_ca_ (a1 a2-f-...+an\a n
J
Dividing co by k, we get CO
(a1
CO
is
Ca
/a2
f1
)
h
k)
(1 I
...
k \I31 1 an
n
I
Now, supposing
d1=(k )a, d2k )a,
..., do
k )a,
we obtain CO
k
_
da
d2 -{- ... - do 1 n
)
(15) 27
Since
( di+d2--...+dn) n
_((a)a+(a)a+-_)ay 1
k
(
= k ca=
n
di+d2-L ...+dn n
a
Ca=-1r
= 1, di+d2+ ... +dn= n.
Suppose
di = 1 +x1, d2 = 1 +x2,
...,
d,, = I +xn. n it follows that From the equality d1 + d2 .+.. + do x1d-x2+. . . -j-xn = 0.
On the basis of Theorem 3 (notice, that a > 1) we have do' = (1 + x1) > 1 +A xi,
d2 = (1 -F x2)" -I+
x2,
a ................
(*)
do = (1 + xn) " > 1-}- a x,t .
Adding these inequalities, we get
d1 +d2 +... } do >n+
a
(Xi+X2+....+x.)=n. (16)
From the inequalities (15) and (16) it follows that
Co _(n)s=1, co>k=ca. 28
It is necessary to note that c13 = k = ca only when the signs of equality occur everywhere in (*), that is when
x1 = x2 = ... = x,, - 0 (Theorem 3). In this case d1 = = d2 = ... = dn, = 1 and, hence, a1 a2 = ... = a = = k. But if the numbers a1, a2, ..., a, are not identical, then Ca > Ca.
Thus Theorem 4 is proved regarding the case when 0 <
If a < P < 0, then 0 < a < I. Reasoning the same way as before, we get in (*) and (16) the opposite signs of ine-
qualities. But since < 0, then from the inequality f3
da -{-d2 +...+do
_<1
it follows that
p 1
rs CO p
k
d1
f3 {
d z } ... - F d na
1
n
that is CO>k=Ca.
Thus, Theorem 4 is proved completely. Further on we shall name the geometric mean by mean
power of the order zero, that is, we shall assume g = co. Notice, that Theorem 4 is applicable in this case as well, since (see Problem 7, Sec. 1.2) ca < g = co, if a < 0, and c5
g-c0, ifI >0.
From the proved theorem it follows, in particular, that C _1 C CO C C1 1 C21
i.e. the harmonic mean does not exceed the geometric mean, the geometric mean in its turn does not exceed the arithme-
tic mean, while the arithmetic mean does not exceed the root-mean-square of positive numbers. For example, if 29
a1= 1, a2 2, a3 = 4, then (aiI+a2I a31 `-1c_
1=`
)
3
=
3
1+ 1
7 =1.7 12
.
co=Vaja2a3=V1.2.4=2, c1= 1+3+4 = 7 = 2.3 ..., 1
3
1+4+16
=Y
2
C2=
J
V7=2.6
3
and therefore
c_1=1.7...<2=co<2, 3... =c1<2.6... =c2. Problem 1. Prove, that x2 + y2 -F- z2 > 12, if x + y + z = 6.
Solution. Since the arithmetic mean does not exceed the root-mean-square, then x -3.1-z
f x2-} y2-}-z2 ) 2 3
l
J
that is x2+y2+z2> (x+3+Z)2 In our problem x2 + y2 + z2' 32 = 12. The sign of equality holds only when x = y = z = 2. Problem 2. Prove, that if x, y, z are positive numbers and x2 + y2 + z2 = 8, then x3+y3+z3> 16
3
Solution. Since c2
1
x2 } y2 } z2 \ 2
( x3+y3+z3 \ 3
I\
3
3
1
In our problem In x3±-y3+z3 1 3 3
30
J
3 8
1
that is
x3+y3+z3j3'
3t/T=16 3
3.
Problem 3. Prove, that for positive numbers al, a2i ., an, the following inequalities are true
a3,
(at
.
.
a2
1f
... +an)a Cna ' (aa+a2 -E- ...
-}-an), a> 1, (17)
(a1 +a2 + .. +an)a>
>na(a"'+a2-}-...+an, 0
(18)
Solution. If a> 1, then C,
( ai-F2--...+an \ a n
1
a1+a2+...+an n
= C1.
The inequality (17) follows easily from this inequality. The inequality (18) is proved in exactly the same way.
In particular, from the inequalities (17) and (18) it follows that (x + y)a < 2a-1(xa + ya), a>
1, x > 0, y > 0,
(x+y)a>2a-1(x"+y"), 0O, y>O. Problem 4. Prove, that if x3 + y3 + z3 = 81, x > 0, y > 0, z > 0, then x+y --f- z<' 9. Solution. Since
(x+y+z)3<, 32(x3+y3+z3) =9.81 =729
(the inequality (17)), then
x+y+z729=9.
31
CHAPTER 2
Uses of Inequalities
The use of inequalities in finding the greatest and the least function values and in calculating limits of some
sequences will be examined in this chapter. Besides that, some important inequalities will be demonstrated here as well.
2.1. The Greatest and the Least Function Values A great deal of practical problems come to various functions. For example, if x, y, z are the lengths of the edges of a box with a cover (a parallelepiped), then the area of the box surface is S - 2xy + 2yz + 2zx,
and its volume is
V _ xyz.
If the material from which the box is made is expensive, then, certainly, it is desirable, with the given volume of the box, to manufacture it with the least consumption of
the material, i.e., so that the area of the box surface should be the least. We gave a simple example of a problem considering the maximum and the minimum functions of a great number of variables. One may encounter similar problems very often and the most celebrated mathematicians always pay considerable attention to working out methods of their solution. Here, we shall solve a number of such problems, making use of the inequalities, studied in the first chapter'. First of all, we shall prove one theorem. 1 Concerning the application of inequalities of the second degree to solving problems for finding the greatest and the least values see the book by I.P. Natanson "Simplest Problems for Calculating the Maximum and Minimum Values", 2nd edition, Gostekhizdat, Moscow, 1952. 32
Theorem 5. If a > 0, a > 1, x y 0,
then the function
the point x= (a)
xa - ax takes the least value in (1-a)(a)a-1.
equal to
a
Proof. The
theorem
very simply for
is proved
the
case when a=2. Indeed, since
x2-ax= (x-
a
a)2
4a2 '
the function has the least value when x = a > 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<6.
Indication. Suppose y = 6 - x. 7. From a square sheet whose side is equal to 2a it is required to make a box without a cover by cutting out a square at each vertex and then bending the obtained edges,
H
Ej
a-2x
20 Fig. 4
so that the box would be produced with the greatest volume (Fig. 4). What should the length of the side of the cut-out squares be?
8. Find the least value of the function x6,+
36
2 -f-
9. Find the least value of the function xs - 8x2 + 5. 10. Find the greatest value of the function
xa-ax when OO, x>O. It. Prove that, when x > 0, the following inequality
is true
3x<8+ 2x. 12. Prove that, when n > 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
rn<1+ 2vn 15. Prove the inequality (1 + al) (1 + a2) . . (1 + an) .
>1+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
lity
vn
vZ v3 20. Find the greatest value of the functions x3
x4+5
Answer.
4 ° 315
;
,
x0-0.6x10.
0.4.
21. At what value of a is the least value of the function
V-+ x2 equal to 2.5? Answer. a = 8. 2.2. The Holder Inequality In Theorem 7, by means of Theorems 5 and 6, the Holder inequality is proved. This inequality will find application in solving problems. Theorem 6. If p > 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)
a2b2
c
-}- d"
,
(')
From these inequalities it follows, that
... + anb,,
a1b1 + a2b2 -}-
di+d2+...
ci+c2+...+cn
+
do
l
/
4
=AB(p-} q)=AB (let us recall that 1
7
p
F-q=1, di d2+...+d9n=1).
Thus, it is proved that the left-hand member of the inequa-
lity (22) does not exceed AB, that is, does not exceed the right-hand member. It is not difficult to mark the case when the sign of equality is valid in (22). Indeed, the sign of equality holds in (21) only when 1
9
1
P
(refer to Theorem 6). Just in the same way, the equality sign will be valid in each line of (*) only when 9
ct=d1p
9
c2=d2P
,
4
,
...,
Cn=dnp
i.e., when p
p
9
C1 = d1, C2 = d2,
P
..., Cn = dn. o
Finally, multiplying these equalities by APB9 we get B9 (Ac1)p = AP (Bd1)9, that is, B9ap = Apbi, a1
b1
=
A BQ
a2
'
b2
=
Ap , B9
an
.,b=
Ap B9
Thus, in (22) the sign of equality is valid if ap a2 an b1b2=...=b;
Note. Taking in the inequality (22) p = 2, q = 2, we get the inequality (19) (refer to Exercise 16): ajb1+ a2b2 + ... -{- anbn
G (a; + az -{- ... + an) (bi + bz + ... + bn). 2.3. The Use of Inequalities for Calculation of Limits In the following problems, the limits of quite complicated -
sequences are calculated by means of previously proved
inequalities.
Problem 1. Prove the inequality
+
(23)
denotes the logarithm from. (1 + In (1 -{with n) base e (see pp. 21-22). n) Solution. Combining the inequalities (8) and (9), we get
(1-f-n)n<e<(1 } n Finding the logarithm of these inequalities with base e, we finally get
nln(1+ n)
n--1 Problem 2. Assuming z1
=1-{ 2 , z2= 2+ 3
-J--
z3=3+4+5+n,
Zy+5-{-1+1+8, 1
find
1
1
lira Zn.
1
n-0 Solution. Substituting n-1 for n in the first memi?e of the inequality (23), we get
n
From this inequality and the second member of the inequality
(23) it follows that In
nn 1 < -1i
(24)
y
Now, using the inequality (24), we write the inequalities In
nn1
nn-1
< n
In n+2 < n+1
nn
1
1 n+2 In n+3 n+2
................
In
2
1
2n
< 2n < In
1
2n2n
Adding them and taking into consideration that the sum of logarithms is equal to the logarithm of the product,
we get (n+1) (n+2) (n+3) ... (2n+1) In < n (n+1) (n+2) ... 2n
n + n+1 + .. . n (n+1) (n-{-2) ... 2n " + 2n < In (n-1) n (n+1) ... (2n-1) 1
1
1
that is In
2nn 1 < n -} n+1 -}- ... -+
2n
n2n1
(25)
Since
2nn 1
= 2 -}- n
,
then
lim In 2nn 1 = lim In (2 + -'
n-00
In 2.
n-.oo
n2 -n
Exactly in the same way from
1
lows,that lim In
n-.oo
n2n 1
2 -}
n
2 -1 it
fol-
In 2.
Thus, the extreme terms of the inequalities (25) have the same limits. I. eice, the mean term has also the sam9 44
limit, that is
+...+2 )=1imZn
1im (n -+- n+1 n-..
Problem 3. Taking x1= 1, x2 = 1- 3 , x3 =
.., xn=
1
+
1
1
1
In 2.
1- 2 + 3 , . .
L(-1) n_
1
1
1
n
calculate lim xn. n-.oo
Solution. We have
x2n=1
+2n1
+
1
1
6-+-
5
1
2n
=(1 } 2 } 3 } 4+ 5+ +-...-+ 2n11+2n 6
-22-4 } 6 } ...+2n
=(1+3+3+4- 5+6+...+2n1 1+2n) -(1+2+6+...+n)= n+1 + n+2 -} ...+2n In the previous problem, we have supposed that
Zn=n+n+1-+-... + 1
1
Therefore, x2n = Zn -
1
1 2n.
But lira zn =1n 2 (refer to the
.
n-w
previous problem). Thus,
Ilm x2n = llm (Zn - n) = In 2.
n-.oo
n-.oo
It is necessary to note also, that x2n+1 =x2n+ 2n+1 , and, hence,
lira x2n+1= lira (x2n + 2n+1 ) -1n 2. n-.oo n-+oo Thus,
lim x, = In 2. n-.oo
Note. The numbers xl = al, x2 = al + a2, x3 = a1 + a2 + c3, . . ., xn = al + a2 +. . . + an are termed 45
partial sums of the series
a1+a2+as+..,an-1--.
.
The series is said to be convergent, if the sequence of its
partial sums has a finite limit. In this case the number S = lim xn is called the sum of the series. n-1 From Problem 3, it follows that the series
1-2+3-4--5-T+---+2n1
1
converges and its sum equals In 2. Problem 4. The series
is called harmonic series. Prove that the harmonic series diverges.
Solution. According to the inequality (23) nn1 .
n>ln
Assuming n =l, 2, 3, ..., n, write n inequalities
1> ln 2
1 2>In2
,
1
3>ln34
,
11 >in nn1 . Adding them, we get xn
-}-
4
2
+
4
+...+
4
n
(n+1)=1n(n+1).
1.2.3 ... n
It follows from this inequality that lira xn>lim In(n-}-1)=oo; n-ao
n-00
hence, the harmonic series diverges. 46
Problem 5. Prove that the series 1
} Za
...
3a
{ na -}-
(26)
converges at any a> 1. Solution. The sequence of partial sums of this series x1 = 1,
x2 = +
2a
x3=1+2a+31
-{X4=1+ V } 4a . ................. 1 xn=1-} 2a -} 3a +...+na1 1
is monotonically increasing, that is
xl <x2 <x3<x4<. . . <xn <. .
.
On the other hand, it is known that monotonically in creasing limited sequence of numbers has a finite limit. Therefore, if we prove that the sequence of numbers xn is, limited, then the convergence of the series (26) will be proved.
as well. Suppose
yen=1-
1
2"-
1
1
1
- na +5a -
V
1
na
... 1
2n-1)a
-
1
(2n)a
Since 1
5a)-...
1
1
1
-
(
_
1
1
1
(2n-1)a)
(2n-2)0'
(2n)a
then (the numbers in each bracket are positive)
Y2n<1.
On the other hand, 1 y2n=1-2a +
_(1-}-2a +
1
3a
1
na
-}-5asa--. 1
1
1
-+-(2n-1)a
-(2n1)a
_
1
_
(2n)a
3a+4+5a+3a+...-+
(2n)a 47
i i = + 6aI +...± (2n)a) -2 (2a+ 4ai 1
1+
3a 4a 1
1
1
2a
1
1
5a
+ 2a -1
`
...
2
2a
+ (2n)0' ) -} ... noc ) 2a+3a 1
}
2a
Since x,, =1 + 2a +
1
(2n -1)a
+ ... + na ,
1
1
then
2
Y2n=x2n-2a xn.
Now, since x2n > xn, y2n < 1, then 2
1>Y2n>xn- 22 xn=
2a-2 2a
xn-
Hence, it follows that 2a 2a-2
xn
that is, the numbers xn are limited when a> 1. Thus, it is proved that the series (26) converges and its sum is a not greater than 2a 2. For example, if a = 2, then
_ 212+32+...+ 1n2<22-2=2, xn+ 22
1
n-+oo
In the course of higher mathematics it is proved that 2
--n2...+n2+...
5=1+"n+
(27)
Exercises 22. Find the sum of the series
S= 1-
1
22
1
+ 32 -
1
`!2 --
... -f- (- 1)
Indication. Use the equality (27).
Answer. S = 48
,2
12 .
_1 1 nz
23. Prove the inequalities na+1 a T1 < I+2a+3a+ ... +nr4<
n+1)
a+
24. Assuming xn = 1 4- 2a -i- 3a -+- ... + na, prove that
lim n . oo
xn
= a+1 1
-(X+ 1
a > 0.
25. Prove the inequality
... + anbncn)3 ... + an) (b1 -E- bz ...
(a ibici -{- a2b2ca -I-
<(a1-f- az
bn) (c1 -1
cz +
... + cn),
if ak, bk, Ch are positive numbers. Indication. Use the inequality (7) and the method given in (22).
26. Assuming xn =
4
n
-}
1
n -+- 1
-{
1
n -{- 2
-}- ... -}
hn,
whe-
re k is a positive integral number, prove that lira xn= Ink. n-. 00
Indication. Use the method of solving Problem 2 of the present section.
2.4. The Use of Inequalities for Approximate Calculation of Quantities At the very beginning of Chapter 1, we have paid attention
to the fact that practical problems require, as a rule, an
approximate calculation of quantities and, as well, an ability to treat such approximately calculated quantities. A more accurate estimation of such quantities will certainly permit to decrease errors in solving problems. In the present section, we are going to return to an approximate calculation of numbers of the form S'n, k = 1/z 4-0866
1
ka
+
(k+1)a
-}-
+ na
,
0 < a < 1,
k
In Sec. 1.1 we have succeeded in finding the number Sn, h
with an accuracy of up to 0.4 for k = 1, n = 1,000,000 and a = 2 (refer to Problem 2). In the same section (see Exercises 2 and 3), for n = 106 and k = 10,000, we were
able to find the number Sn, h already with an accuracy of up to 0.01. The comparison of these two examples shows,
that the indicated method of their solution yields much better results of calculation for greater values of k.
In Sec. 1.4 (Problem 3) we found the integral part of the number Sn, k, for k = 4, n = 106 and a = 1 . Thus, this number was also calculated with an accuracy of up to 0.5. However, we could not find the integral part of the number Sn, i for a = 1 and n = 106 because the method of calculation of such quantities, indicated in Chapter 1, did not permit doing it. In this section, we shall improve the method of calculation of the quantity Sn, 1. This improvement will
make it possible to find similar quantities with a higher degree of accuracy quite easily.
Lemma 1. If x1 > x2 > x3 >
. . .
> xn, then 0
Proof. The number of positive terms in the written
algebraic sum is not less than the number of negative terms. Besides this, the preceding positive terms are greater than the following negative term. This proves that their algebraic sum is positive, A > 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'
<
(2n)1-ani_a
1- a
(28)
Proof. The inequality (28) follows from the inequality (14)
(see Sec. 1.4, Problem 2) when substituting n + I for m,
2n for n and -a for a. 50
Theorem 8. True is the equality 1
1
1
2a
3a
na
2a
1
2-2' 2a
1
2-2a
2a
1
}
- (n+1)a
.
(n 1
{
i
2)a 1
3a-...-(2n)a
(29)
Proof. We have 1
Sn, t= 1 } 2a
1 3a }
... F n } 1
1
(n+1)a
1 1 1 1 + ... + (2n)7 I. + (n+2)a + (2n)a - L (n.11)a
Adding and subtracting from the right-hand member of the equality the number 2 [ 2a -} a2 -}
1
Sn,
=1-r2a-{ +2
L -2a
2a }- ... +
1
-{- 2a
1
}
1
( 2n )
"] ,
_1
3a-4a+...-(2n)"+
a } ...
}
(2n)6'
6
]-
The numbers of the first square brackets have a common factor 2a . Taking it out of the brackets, we get Sn.
2a } 2a + 2a (1
{
2a } 3a
+
(2n)a .
} 2a)
- [ (n+1)a + (n+2)a + ...
{
(2n)a ] . 4*
51
Since in round brackets there is the number Sn,1, then (n+1)a -}-
(n+2)a + ... +
(2n)a -
L1-2+3-. 1
1
. .
1 11
(2n)aJ
-
2a
(2a - 1) S, 1 = 2 2 Sn, 1 Hence, after multiplying by 2a and dividing by 2 - 2a, we get the equality (29). The equality (29) is of interest because it brings the calculation of the quantity Sn,1 to the computation of the quanti1 1 1 t Y S 2n, the quantity ,n, n+1 f 3a
1-
-
-'
2a
(2n)a
The first of these quantities for great it is calculated with a high degree of accuracy by means of the inequality (28). Concerning the second quantity, we know from Lemma 1,
that it is less than zero and greater than -
2a 2--2(x
.
But
if we find the sum of the first four summands of the latter quantity, then the remaining quantity (the error) will be less than zero and greater than - 1 2a 2-2a 5a In the following problems we shall perform the calculation of this quantity with a higher degree of accuracy as well.
Problem 1. Find the sum
A=1 + V2 1
-}-
v3
-{- ... -{-
v106
accurate to 0.002.
Solution. By virtue of Theorem 8 1 -/2 1 A 2-_V2 v106 -{-1 + x/106 -j- 2 + .
-(1-/2 2_v2
__(V
(
+ \ +1)
_/2.106
-...-1 1/3v2.106
1
1
v2 I
+
106+1 11-
.
-VI
2-j-
1
_V2.106
_V3-...-
l
1/2.106
= (V Y+ 1) (B-C), 52
where
B=
1
]-
-V1G6+ 1
C-1 I-
1
v2
+
V106
+
2
i/3 - ... 1
+
-
106
1
v2.106
According to Lemma 2, the number B satisfies the inequalities
2(1/2.101;+1-y106+1)
2 (1/106±1-1!106)-2 (1/2.106-{-1-1/2.100)
_
2
12
-/2.106-{ 1-{- -VT-106;
106-{-1+ -/106
- 1/2 -1
1
1
1/2
66
1/106
1
1,000 ! 3.10 4.
Thus, the middle number will differ from the number B by less than 2.10-4. Calculating the first number and
subtracting from it 2.10-4, we get B = 828.4269 ± A t, I Ar
I<
2.10_4.
Now, proceed to calculating the number C. Let ni be an odd number. Estimate the quantity D
na -m } 1
1
1 1
{
1
2n
-V W, 72
For this reason, it is necessary to notice, that
Vk±I-Vk-1=
2
1/k-71 +1/k-1
and
E
2
2 1/md-1+Vili
+1/m-}-3-- 1/m+1
2
1/m -H- 4 -}-
1/m d- 2
53
2
V2n71-V2n-1 ±V M+V;W-+3-Y-iW -+I
If ;W-+4+
=Vm-
+-Vm+2+ ... -V'2n+1
-Vm-1+V2n-I/2n+1. Thus, the number E is quite easily calculated. Subtracting the quantity D from the quantity E, we get
-
E-D=( Vm+1-2 Vm-1 - Vm 1
2
1
(Vm+2+Vm - Vm-}-1
+ .. .
2 - ( V2n+1-,1/2n--1 - V2 1
Demonstrate, that all the numbers in the brackets are positive and monotonically decreasing. Indeed,
2Vm-(Vm 71 +Vm-1)
2
Vm f 1+Vm-1
Vm (Vm-f-1+V m-1) 2m-2Vm2-1 Vm
=
Vm(Vm+1+Vm-1)(2Vm+V;a I1+Vm-1) 2
Vm (V'n t +V'n-1) (2Vm +V 771+1/m-1) X X (m+Vm2__ 1)
Hence, it is proved, that such numbers are positive and
monotonically decreasing with the increase of m. According to Lemma I
0<E-D< 2
Vm(V 7-}-1+V m-1) (2Vm + V m-}-1+V7 --1) X X (m +Vnt2-1) We shall not make a great mistake, substituting m for the numbers m + I and m - 1 in the denominator. Here, we get
0<E-D< 54
2
1/-m
1
Taking m = 9 we get
O<E-D<8 8 s<0.0006. This proves, that when m=9 and n= 106
E-D=0.0003±A2, A2 1<0.0003, D= E-0.0003 ± 02= j/9 -V 8+ V2.106-
-)/ 2.106 -1-0.0003±A2-0.1710±A2i
i
Now let us return to the quantity C. We have
C = 1 - -2
-1-
+
1
-V4
-V3 1
1
-{- V,3
{
i/4
1/6
V5-
'
1
1
1/5
i/6
{{-
i/
{- D =
+0.1710±A2=
V7-
= 1 - -+0. 1710
- 2 11 + 2) + -
2
4
3y2
2
4
1-
V=3 {
v3 3
/5 -{-
v6
-f- .y7 +
v5- -I y7 5
6
'I_
7
+ 0.1710 t A2.
Thus, for the calculation of the number C with an accuracy of up to 3.10-4 it will be required to find only 5 roots and to produce a number of arithmetic operations. Using the
tables and carrying out necessary calculations, we find
C = 0.6035 ± A 2. Taking into consideration the found quantities B and C, and returning to the quantity A, we get A = (12 + 1) (B - C) _ (12 + 1) (827.8226±A3) _ + 1) .827.8226 ± 2.5A3, where
I 2.5A3 I < 2.5 (1 Al I
I A2 I)<2.5.5.10-4<2.10-3. Thus, the calculation with an accuracy of up to 2.10-3 -I-
will be
A = (V2 + 1) 827.8226 = 1998.539. 55
Problem 2. Calculate the number 1
A=1-r-
1
2 -}
y s+
... +
1
4
1012
with an accuracy of up to unity. Solution. By virtue of Theorem 8 A_
12
2-y 2
1 (
\ y11012+1
1
+
4012+2
V2 /1-
1
y2
+
+.
1
1 2.1012
_
1/1
yj2-1 2 .1012 1.
3
The first term can be easily found and with a high degree of accuracy by means of the inequalities (28). By virtue of these inequalities the first term can he substituted by the number 3
y2
2-y 2
3 (2.1012)4-(1012)4
1_?
4 8
109 (V8-1)
23 4
109.
By virtue of Lemma 1 the sum y2 1 1 i l 2-/2 (1-_V_2 VY is positive and is not greater than the first term. Since the term is less than two, then
---...-21012
3.109-2
Notice that the accuracy of calculating the number A, containing a trillion of addends, is extremely high. The relative
error is less than
100 : 1333333332.3 < 0.0000001 % . 56
Exercises 27. Calculate (with an accuracy of up to unity) the sum 1
...
v22--3 F
{
106.
Answer. 14,999.
28. Show that the equality {
1
...
1
3a
2a
nl-a = 1-a na 1
C
is true, where P,, is an infinitely small quantity, urn Nn = 0, n-.oo
and C
_
2a
T- 2 -
1
1
I
1
1
3M4(Z
...
(-1) n { .,.. n-1
1
l
SOLUTIONS TO EXERCISES 1. Setting
inequalities
the
in
m+1,...n:
21/m+1-21%m<
21/m+2-21/m+1 <
2j/j/m+2<
(p. 9) n = m,
1 <22Vm-1,
/m
1/m+1
V-n
<2j/m+121/m,
<2j/m j-2-2j/m } 1,
+2
21/ n+1-2Vn<
(1)
<2j/n-2j/n-1.
Adding these inequalities we get
2j/ n+1-2j/m<
+ 1/md 1
m
+ ... +
+
+
<21/n-2
_
ti n Vm+2 2. Taking in the inequalities of Exercise I m = 10,000, n = 1,000,000, we obtain 211,000,001 -2 1/10,000< 10,000
+
1
10,001
+
+ 1/1,000,000 <
< 2 j/ 1,000,000- 21/9,999. Since
2 V 1,000,001 > 2 j/ 1,000,000 = 2,000, 2Y10,000=200, 21/9,999 = 1/39,996 > 199.98
(the last inequality can be easily checked, extracting the square root with an accuracy of up to 0.01), then
2,000-200=1,800<
1/10,000 +
+ 1/10,001 +
-L-
1/1,000,000 <
< 2,000 -199.98 = 1800.02. 58
3. Multiplying the inequalities of Exercise 2 by 50, we shall get in our designation 90,000 < 50z < 90,001; hence
[50z] = 90,000.
4. For n = 1, it is obivous, that the inequality is true
2
_ 2.
1
1
1
3.1+1 Assuming now that the inequality is true for n 5 < 3 1 2.4.6 ... 2k+1 2k
k
1
(a)
V3k+1
prove that it is true for n 1
3
5
2.4.6 ...
k + 1, that is, prove that
2k-1 2k+1
1
G i/3k+4
- 2k+2
2k
(h)
Multiplying the inequality (a) by 2k+2 , we get 3 5 2k -1 2k+2 2k +1 1 2 ' 4 ' 6 ... 2k 2k+2< 3k+1 2k+2' What is left is to prove the inequality 1
2k+1
I
1
-j/3k+1 2k-f-2 < i/3k+4
Multiplying it by (2k + 2) V 3k -+1 1/3k + 4 and squaring both parts of the obtained inequality, we get (2k + 1)2 (3k + 4) < (2k + 2)2 (2k + 1), or
12k3 + 28k2 -{- 19k + 4 < 12k3 + 28k2 + 20k + 4. The latter inequality is obvious, since k > 1. This proves that the inequality 1
2n-1 `
3
-T* T "'
2n
'
1
-j/3n+1
is true for all n. 5. Assuming in the inequality of Exercise 4 that n = 50, we get 3 99 < < = 1
T4 ''' 100
1
x/3.50 +1
1
151
1
17144
=
1
12
5* 59
6. Assuming y = 6 - x, x = 6 - y, we shall bring the
problem to finding the greatest value of the function (6 - y) y2 = 6y2 - y3
when 0 < y < 6. Assuming then, that y2 = z, we shall get the function 3
6z - Z2,
whose greatest value (refer to note on p. 34) is equal to 3
H-1
0.5.43=32
32
(-)
7-
and is obtained in the point 1
3_1
Z=
32
=42.
C -2- /
The function 6y2 - y3 takes the greatest value in the
point y = I/ z = 4, and this value equals 32. The function x (6 - x)2 attains the greatest value of 32
in the point x=6-y =6-4 =2.
7. The volume of a box (see Fig. 4, p. 38) equals
V -x(2a-2x)2 =4x(a-x)2, 0<x
V=4 (az-z2). 3
The greatest value of the function az - z2 is obtained in the point 1
3
z
(23 1
2
1
=\3
)2
Therefore, 2a y= /z= 3 , 60
x=a-y=a- 32a = a3 .
Thus, the volume of a box will be the greatest, if the length of the side of the cut-out square is that of the side of the given square. 8. The least value of the function x6 + 8x2 -F 5 equals 5
and is obtained when x = 0.
9. Assuming y = x2, we shall bring the problem to finding the least value of the function
y3 - 8y + 5
for positive values of y. In Theorem 5, we have proved that the least value of the function y3 - 8y is equal to 3
(1-3) (83 )3
3
1=
2
8z
3-
321/6 9
32
The least value of the function y3-8y+5 is equal to 32 9 + 5 = - 3.6 ... 10. Assuming y= xa we get the function
-
.
y-aya (Q y-ya), a >0, a >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
(n+1)n<enn<3nn
Raising both members of the latter inequality to the power of 1 n (n+ 1)' we get n+Vn -{-1 < rn.
then V3 is the
13. Since 1 < j/ 2 = 38 < V9 =
greatest of the numbers 1, Y2, V /-l On the other hand, in the previous problem we have shown that the sequence of the numbers 1Y3, VT' ... , Vn, ... decreases. Hence, V3 is the greatest of the numbers 1, 1/2, V3, ... .
.
. ,
Vn
14. Suppose r n = 1 + (Xn, an > 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<
4
n
an<
,
V2
, ic
an<1--?_ /n=1++ vn
.
Note. Using Newton's binomial, it is easy to check that
rn<+
n
Indeed,
(4+J/-n
)n=1+nf//,?+n(n-1) 2-}-...>1+ 2 n
n
+ n (n -1) 2 2
n = n.
Hence, it follows that
inn<1+Y'rnn 15. When n = 1 and a1 > -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.
+ anbn)2
+ bn) + an) (bi + b2 Notice, that the sign of equality holds only when the polynomial under consideration has a true root, i.e. when (a1 + a2 + at
a2
b 1 = b2
...
an bn
17. Using the inequality (19), we get C1
a1+a2-}-... +an 12 = n
/
a1
( 2
+
1
Vn Vn
< n -{- n + .. . 2
an
1
Vn Vn
I2
(n+n+...+n ) )
=
n
ai - a2 + ... -- an = C2 n 64
2
Hence, it follows that cl < c2 (the arithmetic mean does not exceed the root-mean-square).
18. From the inequality 2=n + 1 + 21/;2 -1+n-1= (y-n+ 1 +
=2n+2Yn2- 1<2n+2Vn2=4n it follows that
Vn +1+Vtn-1 <2Vn, _ 1
1
2Vn <
1/;i-{
1-{ Vn-1
Vn+1-
n-1
Vn-I-1-Vn-i
(V-n+ 1 + n-9) (V;-Fl- n-9)
2
Multiplying by 2, we get 1
Vn
19. Setting in the inequality of Exercise 18n = 2, 3, ... , V n
- <j n+1-Vn-1.
Vn Combining the written inequalities, we get V2+V3+...+
V-n 1
Adding 1 to both parts of the inequality, we finally get 1
V2
-}
- V4 V3
1
{
V5
--
...
1
Vn
< <1/n+1+Vn_
05
Note. It was proved in Sec. 2.1 that 1-F-
The
jr2
+
V3-
-E-...+
V--
>2Yn+1-2V2+1.
+ V n - V2 and 2 Vn + 1 -
numbers
- 2 V2 + 1 differ from each other less than by 0.42. Each of these numbers could be taken for an approximate value of the sum
+
+ r3 + ... +
1/2
Vn-
=zn.
Let us notice without proving, that the number Yn + 1 +
+ V -n - Y Y differs less from the number zn, than the
number 2Vn+1-2V2+1.
20. The function x4+5 takes a negative value when x < 0. Therefore, the greatest value of the function is obtained for positive values of x. Since x3
1
1 x+x3)
+5
5 (5 then the greatest value of the function is reached in the same point in which the function x + x-3 takes the least value. It follows from Problem 5 4 Sec. 2.1 that the least value of this function is equal to =
-3 3-1
(1-4-3)
(
l 3
3
=4(45)4'
The greatest value of the function -+5 is equal to
_
1 3
1
154
_
15
TO- = 20V
a
15
3
4V15
( 151
To find the greatest value of the function x6- 0.6x10, we get y = x6. It is clear that y> 0. The function 10
10
10
y-0.6y6 =0.6 (s y-y6 ) 66
takes the greatest value (see the note on p. 34) equal to 10
6 10
10
6
o
0.6 (6 - 1)
=0.4.
A.snming in this exercise that y=
-
-
a
2
we get
1
2 = y 4 + ay.
l' :z
1
The least value of the iur tion y 4+ay, as it follows from Problem 4 Sec. 2.1. k l- equal to ,.
(14-
('ia;'
4) (4a)5
1
Assuming 4(4a)5=2.5, we get 1
4a = 32,
(4a)5 = 2,
=(1
a = 8.
2z--32-+-4252-} fiz }
-2(22-x[42+62+ ...)= 1
1
1
0+ 22 -} 22
12-x-42 + 52+ 1 + ...
+ -22 +- 32 -x- ... ) _ 1
=Z 1
1
R2
(1+Zz+32+...)=2.
a2
6 -12
(we have used the equality (27)). 23. Since a > 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<
1+a
(n+1)1+a-n1+a
1+a
Write these inequalities for the values n = 1, 2, 3, ... , 1
<1< < 2(<
1+a 21+a-1
1+a
n1+a-(n- 1)1+a
1+a Adding them, we get nl+a T a < 1-{ - 2a - }- 3a -+-
21+a-1 1-f-a 31+a-21+a
1+a
< (n+ 1)1+a-nl+a 1+a +na<(n+1)1+a-1
-
(n+1)l+a 1+a
1-}-a
24. It follows from the inequalities of Exercise 23 1
1--a
<
1+2a
n:
3a } ...+na n1+a
1
-'i
<
that
l+a
n
+
The left-hand member of the latter inequalities is a constant and the right-hand member tends to a limit number
1a ,
equal to 1+a
when n tends to infinity. Hence, the mean member of the inequalities tends to the same limit as well,
that is
,
_ lim 1+2a+ 3a+...+na nl+a n-.oo
25. Let us introduce the designations
B3=b;+b2± ... --bn, C3=C3+C3+... +Cry 68
1
17a
an
ai a2 x1= A , x2= A ,
y1= B
,
y2= B
xn= A
e
bn
yn= B
,
Cl , Z1= C Z2= C , .. .
Zn= CnC
C2
,
On the basis of the inequalities (7) we have xi + A + Z11-
ajb1c1= ABCxiylzl
3
a2b2c2 = ABCx2y2z2 < ABC x2+Y + z2 3
anbncn = ABCxnynzn
+yln
X34
zn
3
Adding the written inequalities we get aibjcj .+ a2b2c2 '_+ _
. .. + anbncn <
+xn + Y i b4+...+y n zi+z2+ ... _
3
3
I
z
ll
).
Taking into consideration the introduced designations, it is easy to calculate that 3
3
3
al+a2--...+4
A3
y3+y2+ ... +yn = 1 , zi+Z3+ ... +Zn = 1. Hence, aibicj +a2b2c2 }
...
}anbncn < ABC ; 3 } 3 {- 3) = ABC.
Raising both members of the inequality to a cube, we finally obtain (albicl -F- a2b2c2 +... + anbnCn)3
A3B3C3 =
_ (al + a2 + ... + an) (bi + b2 + ... + bn) (ci + cz + ... + Cn). 26. Write down the inequalities (24) for different values of n
In nn1 < 3
n n
1
In n+1 < n+1 <_ In n n 1 In
kknl
kn 69
Adding these inequalities, we get (n-{-1) (n+2) ... (kn+ 1)
In
n(n+1)...kn
<
1
1
1
n
+ n-}-1 +
+kn<
n n+1 <1n I n-1 n
kn
1
kn-1
that is
In knn
1
1n (k -{- n <
< n+
1
1
n -F-1
+n-{-2+
1
+kn
CIn
kn
n-1 In (k+
k
n
1
If n tends to infinity, then In (k -E n) tends to In k and k
In (Jr
j
n
1
) tends to the same limit as well. Therefore,
lim (n + n+1 + ...
+kn)=1nk.
n- o0
++10
27. By virtue of Theorem 6 1
_
V2
2-V2
1
(
I y1106 -f- 1
2-V2
+
1
V106+ 2
(1-
+ ... +
+T3
1
1
2.108
1
3p/-2
1
V2.106 /
The second addend is negative but greater than- V3 2-V2 - 1.9. The first addend, according to the inequalities (28), satisfies the inequalities
2 (2.106+1-V106-1) jl2 C 2-V2
1
1
V108-{-1 + 2
70
V 2-V2 108-{-2
< + ... + V2 108 <
(2.106-Y106)
1
= 15,000.
Since the extreme terms of the latter inequalities differ from each other very slightly (less than 0.1), then
15,000-2<1+--+ ... +
10e <15,000. 106
The mean number 14,999 differs from ' s1k less than by 1. k=1
28. By virtue of Theorem 6
2-2 a 2a
- 2 -2
[(n+1)a +(n+2)a + .. .
a
[1-
1
+
- ... -
1
2a -} 3a
(2n)a J
-
]=An-Bn,
(21n)
a
where
2-2" Bn
- 2- 2a [ 12a
1
2a
(2n)]'
(n+2)a
(n+1)a
+
1
3a
- ... -
1
(2n)a
]
.
The number Bn is a partial sum of the series 2a
k- 1
2-2a
(_1)k_1
ka
This series is sign-alternating with monotonically decreasing (by absolute value) terms. Its remainder (by absolute
2a q
value) is not greater than the absolute value of the first term of the remainder, that is, the number . 2 _ 2a
na
Since this number tends to zero when n-} cc, then the series converges and
a
lim Bn = n.oo
2-2a (-1)k ak = C, 2
k_1
71
that is yn = Bn - C is an infinitesimally small value. Now, using the inequalities (28), we get 2a
2-2a [(2n-1)
1-a
-(n+1) 1-a ]
<
[(2n)1-a- n1-a] 2a 2_2a
n1 a
= 1-a
Since the difference between the extreme terms of the inequalities tends to zero when n - - oo, then 6n = An - n1 -a 1--a is an infinitesimally small value. Thus, 1 -I
2a + ... +4-=A-B= n1-a -Sn-G
n1-a
1-a
C
1-a
where Nn = Sn + y is an infinitesimally small value.
TO THE READER
Mir Publishers welcome your comments on the content, translation and design of this book. We would also be pleased to receive any proposals you care to make about our future publications. Our address is: USSR, 129820, Moscow 1-110, GSP Pervy Rizhsky Pereulok, 2 Mir Publishers
Printed in the Union of Soviet
Socialist
Republics
1
The present booklet contains some particularly interesting inequalities playing an important role in various sections of higher mathematics. These inequalities are used for finding
the greatest and the least values as well as for calculating the limits. The booklet contains 63 problems and most of them are provided with detailed solutions. The book is intended for students of senior classes of secondary schools.
llir Publishers Ni -oscow
