Show that n n + 1 3 < 8 1 3 + 2 3 + 3 3 + … + n 3 .

We have to show that n n + 1 3 < 8 1 3 + 2 3 + 3 3 + … + n 3 . The arithmetic mean of the m t h powers of n positive quantities is greater than the m t h power of their arithmetic mean. Let a 1 , a 2 > 0 and a 1 ≠ a 2 and let m be a real number. Then a 1 m + a 2 m 2 > a 1 + a 2 2 m , if m ∈ R - 0 , 1 If m ∈ 0 , 1 then a 1 m + a 2 m 2 < a 1 + a 2 2 m And if m ∈ 0 , 1 then a 1 m + a 2 m 2 = a 1 + a 2 2 m Using the above relation we can derive 1 3 + 2 3 + 3 3 + … … n 3 n > 1 + 2 + 3 + … … + n n 3 ⇒ 8 1 3 + 2 3 + 3 3 + … … + n 3 > n n + 1 3 ⇒ n n + 1 3 < 8 1 3 + 2 3 + 3 3 + … … n 3

If α and β are positive quantities and α > β , show that 1 + 1 α α > 1 + 1 β β . Hence, show that of n > 1 , the value of 1 + 1 n n lies between 2 and 2 · 718 …

If α > β and α , β are positive then to show that 1 + 1 α α > 1 + 1 β β Now, we know that by binomial theorem 1 + 1 α α = 1 + α · 1 α + α ( α - 1 ) 2 ! 1 α 2 + α ( α - 1 ) ( α - 2 ) 3 ! 1 α 3 + … ⇒ 1 + 1 α α = 1 + 1 + 1 - 1 α 1 2 ! + 1 - 1 α 1 - 2 α 1 3 ! . . . . . . ⋯ ( 1 ) Similarly, 1 + 1 β β = 1 + 1 + 1 - 1 β 1 2 ! + 1 - 1 β 1 - 2 β 1 3 ! + . . . . . ⋯ ( 2 ) Now, α > β (given) ⇒ 1 β > 1 α ⇒ - 1 α > - 1 β ⇒ 1 - 1 α > 1 - 1 β Similarly, 1 - 2 α > 1 - 2 β and so on. Hence, after the second, each term of ( 1 ) is greater than the corresponding term of ( 2 ) more over the number of terms in ( 1 ) is greater than the number of terms in ( 2 ) ∴ 1 + 1 α α > 1 + 1 β β Now, in the above result putting α = n and β = 1 , we have 1 + 1 n n > 2 Again putting β = n and α → ∞ , 1 + 1 n n < e < 2 . 718 ∵ lim n → ∞ 1 + 1 n n = e Thus, we conclude that if n > 1 then the value of 1 + 1 n n lies between 2 and 2 . 718 … …

Show that a + b + c + … + k n a + b + c + … … + k < a a b b c c … … . k k

To show that a + b + c + … + k n a + b + c + … … + k < a a b b c c … … . k k . Consider the expression a a b b c c … . . . k k . Suppose that a and b are unequal, if we replace a and b by the two equal quantities a + b 2 , a + b 2 the value of a a b b c c … . . . k k is diminished, since a a b b > a + b 2 a + b . Hence the least value of the expression is when all the quantities a , b , c … k are equal. In this case each of the quantities is equal to a + b + c + … … + k n and the value of a a b b c c … . . , k k then becomes a + b + c … … k n a + b + c + … … + k . Hence when a , b , c , … … k are unequal, a a b b c c … k k > a + b + c + … + k n a + b + c + … + k . Q . E . D .

If n is a positive integer, and x < 1 , show that 1 - x n + 1 n + 1 < 1 - x n n

If x < 1 , we have to show that 1 - x n + 1 n + 1 < 1 - x n n Now, 1 - x n + 1 1 - x n = 1 + x n ( 1 - x ) 1 - x n = 1 + x n 1 - x n ( 1 - x ) = 1 + x n Sum of n terms of a G . P . with first term 1 , common ratio x < 1 = 1 + x n 1 + x + x 2 + … … x n - 1 = 1 + 1 1 x n + 1 x n - 1 + 1 x n - 2 + ⋯ + 1 x Now, since x < 1 , each of the n terms 1 x , 1 x 2 , 1 x 3 , . . . , 1 x n is greater than 1 , and hence their sum is greater than n ∴ 1 - x n + 1 1 - x n < 1 + 1 n ⇒ 1 - x n + 1 1 - x n < n + 1 n ⇒ 1 - x n + 1 n + 1 > 1 - x n n

If x + y + z = 1 , show that the least value of 1 x + 1 y + 1 z is 9 and that ( 1 - x ) ( 1 - y ) ( 1 - z ) > 8 x y z (where x , y , z > 0 ).

Given x + y + z = 1 . Since, x , y , z are positive real numbers, by A . M . - G . M . - H . M . inequality, we have, x + y + z 3 ≥ x y z 3 ≥ 3 1 x + 1 y + 1 z ⇒ 1 3 ≥ 3 1 x + 1 y + 1 z If x = y = z , the equality holds and we get, 1 x + 1 y + 1 z = 9 which is the least value of 1 x + 1 y + 1 z . Now, x + y + z = 1 , so 1 - x = y + z , 1 - y = x + z , 1 - z = x + y Again, A . M . ≥ G . M . , considering only the inequality for a general case, we have, y + z 2 > ( y z ) z + x 2 > ( z x ) x + y 2 > ( x y ) On multiplying these three equations, ( y + z ) ( z + x ) ( x + y ) > 8 x y z ⇒ ( 1 - x ) ( 1 - y ) ( 1 - z ) > 8 x y z Q . E . D

Show that the a b · b a < a + b 2 a + b

To show that a b · b a < a + b 2 a + b Denote ( 1 + x ) 1 - x ( 1 - x ) 1 + x by P ; then ⇒ log P = log 1 + x 1 - x + log 1 - x 1 + x ⇒ log P = 1 - x log 1 + x + 1 + x log 1 - x ⇒ log P = log 1 + x - x log 1 + x + log 1 - x + x log 1 - x = log 1 + x + log 1 - x - x log 1 + x - log 1 - x Since, log 1 + x = x - x 2 2 + x 3 3 - x 4 4 + x 5 5 - . . . , we get log P = x - x 2 2 + x 3 3 - x 4 4 + x 5 5 - . . . + - x - x 2 2 - x 3 3 - x 4 4 - x 5 5 - . . . - x x - x 2 2 + x 3 3 - x 4 4 + x 5 5 - . . . - - x - x 2 2 - x 3 3 - x 4 4 - x 5 5 - . . . = - 2 x 2 2 + x 4 4 + x 6 6 + … … . - 2 x x + x 3 3 + x 5 5 + … … ⇒ log P = - 2 × (positive quantity) - 2 x × (positive quantity) Hence, log P is negative, and therefore P < 1 , i.e., ( 1 + x ) 1 - x · ( 1 - x ) 1 + x < 1 In this result, put x = z u , where u > z ; then 1 + z u 1 - z u 1 - z u 1 + z u < 1 ∴ u + z u u - z u - z u u + z < 1 u ∴ ( u + z ) u - z ( u - z ) u + z < u u - z + u + z Now, put u + z = a , u - z = b , so that u = a + b 2 , therefore a b b a < a + b 2 a + b Q . E . D .

E M B I B E

Higher Algebra>Chapter 19 - Inequalities>EXAMPLES. XIX. a.

H S Hall and S R Knight Solutions for Chapter: Inequalities, Exercise 1: EXAMPLES. XIX. a.

Author:H S Hall & S R Knight

H S Hall Mathematics Solutions for Exercise - H S Hall and S R Knight Solutions for Chapter: Inequalities, Exercise 1: EXAMPLES. XIX. a.

Attempt the practice questions on Chapter 19: Inequalities, Exercise 1: EXAMPLES. XIX. a. with hints and solutions to strengthen your understanding. Higher Algebra solutions are prepared by Experienced Embibe Experts.

Questions from H S Hall and S R Knight Solutions for Chapter: Inequalities, Exercise 1: EXAMPLES. XIX. a. with Hints & Solutions

HARD

JEE Main

IMPORTANT

Higher Algebra>Chapter 19 - Inequalities>EXAMPLES. XIX. a.>Q 3

Show that $n {(n + 1)}^{3} < 8 (1^{3} + 2^{3} + 3^{3} + \dots + n^{3})$ .

HARD

JEE Main

IMPORTANT

Higher Algebra>Chapter 19 - Inequalities>EXAMPLES. XIX. a.>Q 6

If $α$ and $β$ are positive quantities and $α > β,$ show that ${(1 + \frac{1}{α})}^{α} > {(1 + \frac{1}{β})}^{β}$ .

Hence, show that of $n > 1,$ the value of ${(1 + \frac{1}{n})}^{n}$ lies between $2$ and $2 \cdot 718 \dots$

HARD

JEE Main

IMPORTANT

Higher Algebra>Chapter 19 - Inequalities>EXAMPLES. XIX. a.>Q 9

Show that
${(\frac{a + b + c + \dots + k}{n})}^{a + b + c + \dots \dots + k} < a^{a} b^{b} c^{c} \dots \dots . k^{k}$

HARD

JEE Main

IMPORTANT

Higher Algebra>Chapter 19 - Inequalities>EXAMPLES. XIX. a.>Q 12

If $n$ is a positive integer, and $x < 1,$ show that $\frac{1 - x^{n + 1}}{n + 1} < \frac{1 - x^{n}}{n}$

HARD

JEE Main

IMPORTANT

Higher Algebra>Chapter 19 - Inequalities>EXAMPLES. XIX. a.>Q 15

Find the maximum value of $x^{3} {(4 a - x)}^{5}$ if $x$ is positive and less than $4 a$ and find the maximum value of $x^{\frac{1}{2}} {(1 - x)}^{\frac{1}{3}}$ , when $x$ is a proper fraction.

HARD

JEE Main

IMPORTANT

Higher Algebra>Chapter 19 - Inequalities>EXAMPLES. XIX. a.>Q 18

If $x + y + z = 1,$ show that the least value of $\frac{1}{x} + \frac{1}{y} + \frac{1}{z}$ is $9$ and that $(1 - x) (1 - y) (1 - z) > 8 x y z$ (where $x, y, z > 0$ ).

HARD

JEE Main

IMPORTANT

Higher Algebra>Chapter 19 - Inequalities>EXAMPLES. XIX. a.>Q 21

Show that the expressions
$a (a - b) (a - c) + b (b - c) (b - a) + c (c - a) (c - b)$ and $a^{2} (a - b) (a - c) + b^{2} (b - c) (b - a) + c^{2} (c - a) (c - b)$ are both positive (where $a, b, c > 0$ ).