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20

2Inequalities

1. Inequality exists only between two real numbers (not complex numbers). 2. If a be any real number then one and only one of there hold.

0,a 0, 0a a

3. If , 0a b then 0, 0.a b ab

4. (i) a b if 0a b (ii) a b if b a (iii) a b if either a b or a b (iv) a b if either a b or a b 5. In a given inequality, terms/coefficients from one side to other side can be transferred as in the case

of an equality. 6. On can add/subtract the same real number on both sides of an inequality, the direction of inequality

does not change. 7. Two is equalities with same direction can be added (always) and multiplied (if both sides of the

inequality are positive). But they can never be subtracted or divided. 8. Both sides of an inequality can be multiplied by same positive quantity without changing the

direction of inequality. 9. The direction of inequality changes if it is multiplied both sides by a negative number. 10. If a b then ac bc if , , 0.a b c

11. If a b then ( ) ( )f a f b , if f(x) is an increasing function of x and also if ,a b then ( ) ( )f a f b if ( )f x is decreasing function of x.

12. For a closed convex polygon in XY-plane, any linear function of x and y i.e., z ax by defined over such a convex polygon will have maximum and minimum value only at the vertices of the polygon.

13. If a c and b c then b c a or .b a Also if a b and b c then .a c 14. Functions ( , , ), ( , , )f a b c g a b c and ( , , )h a b c can not be positive simultaneously if 0f g h

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21

15. Jensen’s Inequality : If ( ) ( )2 2

x y f x f yf for ( ), , [ , ]2

x yx y a b

then ( )f x is positive and curves towards x-axis at least in the given domain.

If ( ) ( )2 2

x y f x f yf for , , [ , ]2

x yx y a b

then the function curves away from x-axis. 16. Method of induction is very useful in proving result of inequality involving only natural number.

17. If 1 2 2 3 1, , ..., n na a a a a a then 1 .na a

18. If 1 1 2 2, , ..., n na b a b a b then 1 2 3 1 2... ...n na a a a b b b and equality sign holds iff

1 1 2 2, , ..., .n na b a b a b

19. (i) If 0a b then 1 1 .b a

(ii) If 0a b and 0c d then a bd c and .ac bd

(iii) If 0a b and 0a d then .a bd d The equality sign holds iff a b and .c d

20. (i) Let 0, , 0a b p q and let 1/qa and 1/qb denote positive qth roots of a and b respectively then

/ /p q p qa b and / /p q p qb a

(ii) Let 0, ,a b p z e a non negative integer and q a positive integer and 1/ 1/,q qa b denote qth roots of a and b respectively then

/ /p q p qa b and / /p q p qb a The equality sign holds of and only if a b or 0.p

21. (i) For two positive numbers a and b, let arithmetic denotes A. mean, G denotes geometric mean and ‘H’ denotes Halmount mean then .A G H

The equality sign holds if and only if .a b (ii) If A, G and H be respectively the arithmetic mean, the geometric mean and Harmonic mean

of n positive integers 1 2, , ..., na a a then .A G H The equality sign holds if and only if

1 2 .... .na a a

22. (i) If x is real and 2 0Ax Bx C 2 4 0.B AC

(ii) If 0A and x is real then 2 0Ax Bx C

24 0AC B 23. Triangle Inequality : | | | | | | | | | |a b a b a b . 24. | | | |.i iii a a

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22

25. The lengths a, b, c can represent the sides of a triangle if and only if , , .a b c b c a a c b

26. Weierstras’s Inequality : For positive number 1 2 3, , , ..., na a a a

1 2 3 1 2 3(1 )(1 )(1 ) ... (1 ) 1 ...n na a a a a a c a

If ia are fractions (less than one) then

1 2 1 2 3(1 )(1 ) ... (1 ) 1 ( ... )n na a a a a a a

27. Cauchy Schwartz Inequality : 2 2 2 2 2 2 21 1 2 2 1 2 1 2( ... ) ( ... )( ... ).n n n na b a b a b a a a b b b

28. Tchebychev Inequality : If 1 2 3 ... nx x x x and 1 2 3 ... ny y y y or 1 2 3 ... nx x x x and

1 2 3 ... .ny y y y then

1 1 2 2 1 2 1 2... ... ...n n n nx y x y x y x x x y y yn n n

If one of the sequences is increasing and other is decreasing then the direction of the inequality changes.

29. If 1 2{ , , ..., }na a a a are positive numbers, { }ib b are various permutations of ia then

2

1 1

n ni i i

i ia a b

i.e., 2 2 3 2

1 2 3 1 2 2 3 1 3 1 .n na a a a a a a a a a a a

30. The product k l m na b c d when a b c d Z will be maximum when k l m na b c d

k l m n

attains maximum. Also using AM GM

.k l m n

k l m n k l m nza b c d k l m nk l m n

31. Holder’s Inequality : 1 1 2 2 1 2 1 2( ... ) ( ... ) ( ... )p p q qpq p q q pn n n na b a b a b a a a b b b

Where 1 1 1,P q ia and ib are non negative real numbers.

Example 1. For any three positive real numbers a, b, and c, show that 2 2 2 .a b c ab bc ca Solution : We know that

2a bab

by squaring we have

2 2 2

4a b abab

2 22

4 4ab a bab

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23

2 22

4 4ab a b

2 2

2a bab

Similarly 2 2 2 2

,2 2b c c abc ca

Adding these three relations, we get

2 2 2 2 2 2

2 2 2a b c b c aab bc ca

2 2 22( )

2a b cab bc ca

. Hence proved.

Example 2. Show that 4 4 4 2 2 2 2 2 2.x y z x y x z y z

Solution: Let the three numbers be 4 4,x y and 4.z

Applying A.M.,G.M. inequality to 4x and 4.y 4 4

4 42

x y x y

4 4 2 22x y x y …(1)

Similarly, 4 4 2 22x z x z …(2)

And 4 4 2 22y z y z …(3)

Adding (1), (2) and (3), we get

4 4 4 2 2 2 2 2 2x y z x y x z y z . Hence proved.

Example 3. If 2 2 2 27x y z show that 3 3 3 81.x y z

Solution : Applying Cauchy-Schwartz inequality, to 2 sets

3/2 3/2 3/2( , , )x y z and 1/2 1/2 1/2( , , ),x y z

We have 2 2 2 2 3 3 3( ) ( )( )x y z x y z x y z …(1)

Again applying C-S inequality, to ( , , )x y z and (1,1,1).

We have 2 2 2 2( ) 3( )x y z x y z …(2)

Squaring both sides of (1) we have 2 2 2 4 3 3 3 2 2( ) ( ) ( )x y z x y z x y z

On using (2), the above inequality yields,

2 2 2 4 3 3 3 2 2 2 2( ) 3( ) ( )x y z x y z x y z …(3)

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24

Since 2 2 2 27,x y z from (3), we get

3 3 3 2 2( ) (81)x y z

Taking +ve square root,

3 3 3 81x y z Hence Proved.

Example 4. Show that ( )( ) 4 ;px qy pq xy pxqy given p, q, x, y are +ve.

Solution : Given p, q, x, y are +ve , , ,px qy pq xy are +ve numbers

Applying A.M., G.M. inequality to px and qy.

2px qy pxqy

2px qy pxqy …(1)

Again, Applying A.M.– G.M. inequality to pq and xy

2pq xy pxqy

2pq xy pxqy …(2)

Multiplying (1) and (2) we get ( )( ) 4 .px qy pq xy pxqy

Example 5. Show that 3 3 3 3 .a b c abc

Solution : If 3 3,a b and 3c we get

Applying A.M. – G.M. inequality. 3 3 3

3 3 3 33

a b c a b c

3 3 3 3 .a b c abc Hence Proved

Example 6. If a, b, c are unequal and positive show that 2 2 2 .a b c ab bc ca Solution : a, b, c are unequal and positive.

Applying A.M. – G.M. inequality to 2a and 2.b 2 2

2 22

a b a b

2 2 2a b ab …(1)

Similarly, 2 2 2b c bc …(2)

2 2 2c a ca …(3)

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25

Add to get 2 2 2a b c ab bc ca

Example 7. If a be a positive number, prove that 1 2.a a

Solution : Let two positive numbers be a and 1 .a

By A.M., G.M. Inequality 1

12

a a a a

1 2.a a H.P.

Example 8. Prove that : 2

4121

xx

or 2

21 2x

x

or 22

1 2.xx

Solution : Let the 2 positive numbers be 2x and 21 .x

By A.M. – G.M. Inequality,

22 2

2

11

2

xx x

x

22

1 2xx

4

21 2x

x

2

41 .21

xx

Hence Proved

Example 9. Proved that tan cot 2 .

Solution : Applying A.M. – G.M. inequality to 1tan tan

1tan 1tan tan2 tan

(By A.M. G.M.)

tan cot 2

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26

Example 10. Which is bigger 1417 or 1131 .

Solution : Now 14 1417 16

14 4 1416 (2 )

14 56 5517 2 2

55 5 11 112 (2 ) (32)

14 11 1117 32 31

Thus 14 1117 31

Example 11. Sow that 99 99 99 99 991 2 3 4 5 is divisible by 5.

Solution : 99 99 99 99 99 99 99 99 99 991 2 3 4 5 (1 4 ) (2 3 ) ... 5 . …(1)

99 99 98 97 96 2 981 4 (1 4)(1 1 .4 1 .4 ... 4 ) 5 p

Similarly, 99 992 3 5 q

and above 995 5 q

thus the expression is divisible (1) is divisible by 5 Example 12. For positive a, b, c prove that.

3 3 3 ( )a b b c c a abc a b c

Solution : Let the two sets of numbers be { , , }A a b c

2 2 2{ , , }B a b b c c a

By Tchebycheff inequality 2 2 2 2 2 2( ) ( ) ( ) ( )

3 3 3a b b b c c c a a b c a b b c c aa

3 3 3 2 2 23( ) ( )( )a b b c c a a b c a b b c c a …(1)

by A.G.–G.M. Inequality

2 2 2

3 3 3 33

a b b c c a a b c

2 2 2 3a b b c c a abc …(2) Hence from (1) and (2), we get

3 3 3 ( )( ).a b b c c a a b c abc

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Example 13. For positive a, b, c prove that

32

a b cb c c a a b

Solution : As a, b, c are positive. We may assume that a b b c c a

Hence 1 1 1a b b c c a

Let { , , }A a b b c c a

1 1 1, , .B a b b c c a

Now apply Thebycheff’s inequality

3 {( ) ( ) ( )}a b b c c a a b b c c aa b b c c a

1 1 1( ) ( ) ( )a b b c c a

1 1 13 3 2( )a b c a b b c c a

92

a b c a b c a b ca b b c c a

9 32c a b

a b b c c a

32

a b cb c c a a b

Example 14. For positive a, b, c such that 1abc show that 1b c c a a ba b c . Solution : From the given expression

( ) ( ) ( )b c c a a ba b c 1 c ab c a ba cac

[ 1]abc

( ) ( )b c c a c a a ba a c c

b c c a

c a a bac

.b a

c bac

We can assume that a b c as the equation is symmetric. So the numbers are positive integers raised to positive powers. So the denominator is greater than the numerator. Hence the above expression is 1.

i.e., 1b c c a a ba b c

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Example 15. Detemine all real number satisfying the inequality,

1 log(2 1) log ( 9) 1.2 x x

Solution : 1 log(2 1) log ( 9) 12 x x

log (2 1) log 9 1x x

1/2log[(2 1)( 9)] log10x x

1/2 2 2[(2 1)( 9)] (10)x x

22 18 9 100x x x

22 18 91 0x x x

22 19 91 0x x

7 132 x

But for log to be defined 2 1 0x and 9 0x

12x and x > 9.

i.e., 9 13.x Hence the required value 9 13x Ans.

Example 16. If a, b, c, d are positive real numbers then show that 1 1 1 1( ) 16a b c d a b c d

.

Solution : Without loss of generality, a b c d

1 1 1 1 .a b c d

Applying Tchebycheff’s inequality

1 1 1 1

4 4 4

a b c da b c d a b c d a b c d

1 1 1 1( ) 4 4a b c d a b c d

or 1 1 1 1( ) 16.a b c d a b c d

Example 17. If , , ,a b c d are +ve, prove that

5 5 5 5 ( ).a b c d abcd a b c d

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Solution : We choose two sets 4 4 4 4( , , , ) and ( , , , )a b c d a b c d

Applying inequality 5 5 5 5 4 4 4 44( ) ( )( )a b c d a b c d a b c d

Applying A.M. – G.M. inequality to 4 4 4 4, , ,a b c d

4 4 4 4

4 4 4 4 44

a b c d a b c d

4 4 4 4 4a b c d abcd

5 5 5 5 ( ).a b c d abcd a b c d

Example 18. Show that ( 1) 2.4.6... 2 .nn n

Solution : Consider the numbers 2.4.6... 2 .n

A.M. G.M.

2 4 6 ... 2 2 4 6 ... 2nn nn

[4 2 2]2 2 4 6 ... 2n

n nnn

( 1) 2 4 6 ... 2nn n

( 1) 2.4. 6... 2 .nn n

Example 19. For positive numbers , ,x y z Show that

3( ) 27( )( )( )x y z y z x z x y x y z

Solution : Consider the numbers : ,y z x z x y and x y z

3 ( )( )( )3x y z y z x z x y x y z

3( ) ( )( )( )27

x y z y z x z x y x y z

3( ) 27( )( )( )x y z y z x z x y x y z

Example 20. Show that 11 2

2 3 1... n n

n

x xx x nx x x x if 1 2, , , nx x x are positive.

Solution : Consider the numbers:

11 2

2 3, , ..., ,n n

n

x xx xx x x x

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30

A.M. A.M.

11 22 3 1 3 11 2

2 3 4

..., , , ..., .

n n

n n n

n

x xx xx x x x x x xx x

n x x x x x

3 11 2

2 3 4 1... .n n

n

x x xx x nx x x x x

Example 21. Show that 1.3, ..., (2 1).nn n

Solution : Consider the numbers 1, 3, 5, ..., (2 1).n

1 3 5 ... (2 1) 1.3.5 ... (2 1)nn nn

[2 2 2]2 1.3.5 ... (2 1)n

n nnn

( ) 1, 3, 5, ..., (2 1)nn n

Example 22. Find the minimum value of 2x y subject to the condition 8,xy x and .y R

Solution : Applying A.M. – G.M. inequality

2 2 .2x y x y

2 2.82x y

2 42x y

2 8x y

Minimum value of 2 8.x y

Example 23. Let , , ,a b c d be real numbers such that .a b c d

Prove that 2( ) 8( )a b c d ac bd

Solution : Consider the quadratic polynomial with real coefficients

( ) ( )( ) ( )( )f x x a x c x b x d or 2( ) 2 ( ) ( )f x x a b c d x ac bd

since a b < ,c d ( ) 0, ( ) 0, ( ) 0, ( ) 0f a f b f c f d

Hence ( ) 0f x has real root between a and b and also between c and d (by Descarte’s rule of signs) i.e., the quadratic equation ( ) 0f x has roots which are real and distinct.

Discriminant is positive

i.e., 2( ) 4 2( )a b c d ac bd

i.e., 2( ) 8( )a b c d ac cd

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Example 24. Without using tables, prove that

2 5

1 1 2log log

Solution : Let 2log a and 5log b

2a and and 5b

1/2 a and 1/5 b

1 1

2 5 a b

i.e., 1 1

10 a b

But 210 . (Since 227

)

1 1

2 a b

1 12 a b

or 1 1 2a b

Show that 2

1 1 2log log 2

Example 25. If a, b, c are three numbers 0,

such that 1,a b c prove that

13ab bc ca

Solution : Now

2 2

2 2

2 2

222

a b abb c bcc a ca

2 2 2a b c ab bc ca …(1)

2( ) 2( )a b c ab bc ca ab bc ca

i.e., 1 3( )ab bc ca

1 .3ab bc ca

Example 26. Let a, b, c be real numbers with 0 1, 0 1, 0 1a b c and 2.a b c

Prove that 81 1 1a b c

a b c

.

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Solution : If , ,x y z are positive, then

2 , 2x y xy y z yz 2z x zx

( )( )( ) 8x y y z z x xyz …(1)

Thus 8(1 )(1 )(1 ) [(1 1 )(1 1 )(1 1 )]a b c a b b c c a

8(1 )(1 )(1 ) (2 ) (2 ) (2 )a b c a b b c c a

cab

. . 81 1 1a b c

a b c

Example 27. Show that, for a triangle with radii of circum circle and incircle equal to R and r respectively, the inequality 2R r hols.

Solution : 4abcR

and r s

2 4. ( )( )( )4R abcs abcsr s s a s b s c

2( )( )( )

abcb c a c a b a b c

…(1)

Now, applying A.M. – G.M. inequality. ( ) ( ) ( )( )2b c a c a b b c a c a b

i.e., ( )( )c b c a c a b …(2)

similarly, ( )( )b c b a a b c …(3)

( )( )a c a b a b c …(4)

( )( )( )abc b c a c a b a b c …(5)

1( )( )( )abc

b c a c a b a b c

…(6)

2 2( )( )( )abc

b c a c a b c b c

…(7)

i.e., 2Rr or 2R r …(8)

Example 28. Prove that 1 1 1 11 ... 1 .1001 1002 3001 3

Solution : Let 1 1 1 1... ...1001 3001 1001 3001s

1 1 12000 2002 2001

For any n, 2 24002 2001 0n n and hence 2(4002 ) 2001n n

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33

2 2

1000 times

1 1 14002 ... 20012001 2001s

21000 14002 20012001

2000 12001

1

Again, rate, terms hundred at a time

1 1 1 1 1... 100 ...1001 3001 1001 1101 2901S

1 1 1100 ...1000 1100 2900

1 1 1..10 11 29

1 1 1 15 10 15 20 25

1 1 1 1 77 112 3 4 5 60 3

Hence 11 13s

Example 29. If 1abcd show that (1 )(1 )(1 )(1 ) 16.a b c d

We expand (1 )1 )(1 )(1 )a b c d and collect the terms in pairs such that the product of the terms in each pair is 1. Thus,

4(1 )(1 )(1 )(1 ) (1 ) ( )a b c d abcd a bcd

3( )ab cd

where the integer written under the (symbol) denotes the number of terms governed by

Now, each of the terms (1 ), ( )abcd a bcd and ( )ab cd

is of the firm 1x x (with 0)x and so it is 2 .

Since there are 1 4 3 8 such terms, we have the sum 1 1 11 1 1 8x y z

4( ) ( ) ( ) ( ) ( )a bcd a bcd b acd c abd d abc

3( ) ( ) ( ) ( )ab cd ab cd ac bd bc ad

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34

Example 30. Show that 0x then 3 23 6 4 0.x x

Solution : Since 3 3 33 4 2 4,x x x

applying A.M. – G.M. inequality, 33 3 3 3 2 22 4 3. 2 .4 3.2 6x x x x x x

Thus 3 23 6 4 0x x Example 31. Show that if x, y, z are non-negative reals, such that 1.x y z

Solution : 1,x y z then 1 1 11 1 1 8x y z

1 1 ( )1 x x y z x y zx x x x

also 2y z yz

( )( )( ) 8y z z x x y yz zx xy

8xyz

. . 8y z z x x yx y z

Example 32. If a, b, c, d are positive real numbers. Show that

1 1 1 1( ) 16a b c d a b c d

.

Solution : 1/4( )4a b c d abcd

and 1/41 1 1 1

1 1 1 1. .4a b c d

a b c d

Multiplying LHS and RHS respectively.

1 1 1 1( ) 16.a b c d a b c d

Example 33. If a, b, c are positive real numbers, prove that, 2 2 2 2 2 2

( )b c c a a b a b cb a c a a b

Solution : Now 2 2 21 ( )2b c b c …(i)

2 2 2

2 2b c b c

2 2 1 ( )2

b c b cb c

…(ii)

Similarly 2 2 1 ( )2

c a c ac a

and 2 2 1 ( )2

a b a ba b

…(iii)

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Adding 2 2 2 2 2 2 1 (2 2 2 )2

b c c a a b a b cb c c a a b

( ).a b c

Example 34. Show that 2 31 ( !) .2

nn nn n

Solution : Consider the unequal positive numbers 3 3 3 31 , 2 , 3 , ..., .n

.e., 3 3 3 3

3 3 3 3 1/1 2 3 ... (1 .2 .3 ... ) nn nn

(A.M. G.M.)

i.e., 2 2

3 1/3( 1) [( !) ]4

n n nn

Raising both sides to powers n,

3( 1) ( !)4

nn nn n

i.e., 2 31 ( !)2

nn nn n

Example 35. If a, b, c are positive and 1a b c . Show that 1 1 1 9.a b c

Solution : On dividing by ( )a b c successively, by , ,a b c we get

11 b ca a a …(i)

11a cb b b …(ii)

11a bc c c …(iii)

Adding, 1 1 1 3 a b b c a ca b c b a c b c a

3 2 2 2 9

Example 36. Show that 201 1log 3 .3 2

Solution : 9 20

20 20log 9 log 20

220log (3) 1

i.e., 202log 3 1

201log 3 2 …(i)

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36

27 20

20 20log 27 log 20

i.e., 320log (3) 1

20 2013log 3 1 log 3 3 …(ii)

Thus 201 1log 3 .3 2

Example 37. In ,ABC Show that 3 2.2a b c

b c c a a b

Solution : Now 1/3( ) ( ) ( ) {( )( )( )}3a b b c c a a b b c c a

…(i)

1/31 1 1

1 1 13

a b b c c aa b b c c a

…(ii)

Multiplying L.H.S. 1 1 1 9( ) 2a b c a b b c c a

…(iii)

i.e., 9 332 2a b c

b c c a a b

…(iv)

Also b c a

a a ab c a b c

i.e., 2a ab c a b c

2( ) 2a b c a b cb c c a a b a b c

Thus 3 22

a b cb c c a a b

Equality occurs when .a b c

Example 38. If a, b, c are sides of triangle show that 1 1 1 1a b cb c c a a b

a b c

.

Solution : Since , ,a b c are the sides of a triangle,

0, 0, 0a b c b c a c a b …(i)

Thus 1 ,1 ,1b c c a a ba b c

are all positive …(ii)

Take 1 ,1 ................. ‘ ’ timesb c b c aa a

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1 ,1 ................. ‘ ’ timesc a c a bb b

1 ,1 ................. ‘ ’ timesa b a b ac c

and apply A.M. – G.M. inequality

1 1 1b c c a a ba b ca b c

1

1 1 1a b c c b cb c c a a b

a b c

LHS of the inequality 1 i.e., 1 R.H.S.

Thus 1 1 1 1a b cb c c a a b

a b c

.

Example 39. If a, b, c, d are four non-negative real numbers and 1,a b c d show hat 1 .4ab bc cd

Solution : 2( ) 4( )a b c d ab bc cd

2 2 2 2 2 2 2 2 2 2a b c d ab bc cd ac ad bd

2 2 2 22 2 2 2 2 2a ab b c d cd bc ac ad bd

2 2( ) ( ) 2( )( ) 4a b c d a b c d ad

2[( ) ( )] 4 0a b c d ad ( , , , 0)a b c d

1 4( ) 0ab bc cd

4( ) 1ab bc cd

1( ) 4ab bc cd

Aliter : the above problem can be solved by using A.M. – G.M. inequality we know that ( ) ( ) 1.a c b d

2 ( )( ) ( ) ( )a c b d a c b d

2 ( )( ) 1a c b d

4( )( ) 1a c b d

14ab ad bc cd

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1 .4ab bc cd ad

1 .4ab bc cd ( 0)ad

Example 40. If , , ,a b c d are positive, then prove that

3 3 3 3 3 3 3 3 2( )( ) 16( ) .a b b c c d d a ab bc cd da abcd

Solution : Applying Cauchy Schwarz inequality 3 2 3 2 3 2 3 2

1 2 3 4, , ,a b a b c a c d a d a a

and 3 2 3 2 3 2 3 21 2 3 4, , , ,ab b bc b cd b da b

we get 2 2 2 2 2 2 2 21 1 2 2 3 3 4 4, , , .a b a b a b b c a b c a a b d a

Now applying A.M. – G.M. inequality, we get

242 2 2 2 2 2 2 2 2 4 4 4 4 2 2 2 2( ) 4 16a b b c c d d a a b c d a b c d

and hence 3 3 3 3 3 3 3 3 2( )( ) 16( ) .a b b c c d d a ab bc cd da abcd

Example 41. Given that 2 2 2 8,x y z prove that

3 3 3 216 .3x y z

Solution : Applying Cauchy Schwartz inequality with

Let 3/2 3/2 3/2, ,x y z and 1/2 1/2 1/2, ,x y z

we have 2 2 2 2 3 3 3( ) ( )( )x y z x y z x y z

Again 1 1 1x y z x y z

so 2 2 2 2 2 2 2( ) ( )(1 1 1 )x y z x y z

( ) 3 8x y z

and hence 2 2 2 2

3 3 3 ( ) 64( ) ( ) 3. 8x y zx y z x y z

3 3 3 64 2.4 3x y z

3 3 3 216 .3x y z

Example 42. If 3 3 3 3 10,w x y z show that

4 4 4 4 3 2500w x y z

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Solution : Applying Cauchy Schwarz inequality for 2 2 2 2, , ,w x y z and , , , ,w x y z we get

3 3 3 3 2 4 4 4 4 2 2 2 2( ) ( )( )w x y z w x y z w x y z …(1)

Again applying Cauchy Schwarz inequality with 2 2 2 2, , ,w x y z and 1, 1, 1, 1, we get

2 2 2 2 2 4 4 4 4 4( ) ( )w x y z w x y z

2 2 2 2 4 4 4 4 2( ) ( )w x y z w x y z …(2)

3 3 3 3 2

4 4 4 42 2 2 2

( )( ) ,( )w x y zw x y zw x y z

by Eq. (1)

2 3 3 3 2

4 4 4 4 1/ 2( ) ,

2( )w x y zw x y z

by (2)

4 4 4 4 3/2 100( ) 502w x y z

4 4 4 4 2/350w x y z or 3 2500

Example 43. If 2( 1) (5 1)x x and 2( 1) (7 3),x x find the integral values of x.

Solution : We have 2( 1) 5 1x x

2( 1) (5 1) 0x x

2 3 2 0x x ( 2)( 1) 0x x

2x …(1) and 1x …(2)

Again 2( 1) (7 3)x x

2 5 4 0x x ( 4)( 1) 0x x

1 4x …(3) Again Eq. (2) and Eq. (3) cannot satisfy one another. Hence, we should consider Eq. (1) and Eq. (3) from which we get

2x and 4x so 3.x Example 44. Prove that the polynomial

9999 8888 7777 1111... 1x x x x is divisible by 9 8 7 ... 1.x x x x

Solution : Let 9999 8888 7777 1111... 1P x x x x

and 9 8 7 ... 1Q x x x x

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9 9990 8 8880( 1) ( 1)P Q x x x x 7 7770 1110( 1) ... ( 1)x x x x

9 10 999 8 10 888[( ) 1] [( ) 1]x x x x 7 10 777 10 111[( ) 1] ... [( ) 1]x x x x …(1)

But 10( ) 1nx is divisible by 10 1x for all 1n

RHS of Eq. (1) is divisible by 10 1.x

P Q is divisible by 10 1x and hence divisible by 9 8 ... 1.x x

Example 45. Given that the equation 4 3 2 0x px qx rx s has four positive roots, prove that (i)

16 0,pr s (ii) 2 36 0.q s

Solution : Let , , , be he four positive roots of the given equation. Then,

p …(1)

g q …(2)

r …(3)

s …(4)

(i) Using A.M. – G.M. inequality in Eq. (1) and Eq. (3), we get

3 3 3 344.4 4

s

.4 4p r s

16pr s or 16 0pr s

(ii) Applying A.M. – G.M. inequality in Eq. (2), we get

3 3 3 366q s

2 36q s or 2 36 0.q s

Example 46. If a, b and c are positive real numbers such that 1,a b c prove that (1 )(1 )(1 ) 8(1 )(1 )(1 ).a b c a b c

Solution : We know 1a b c 1b c a and 1 1 1 ( ) (1 ) (1 )a b c b c

and since 1a b c where a, b and c are positive real numbers, so 1 b and 1 c are positive.

Applying A.M. – G.M. inequality, we get

1 (1 ) (1 ) 2 (1 )(1 )a b c b c …(1)

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41

1 (1 ) (1 ) 2 (1 )(1 )b a c a c …(2)

and 1 (1 )(1 ) 2 (1 )(1 )c b a b a …(3)

Multiplying Eq. (1), Eq. (2) and Eq. (3), we get (1 )(1 )(1 ) 8(1 )(1 )(1 ).a b c a b c

Example 47. If a, b and c are the sides of a triangle and 2,a b c then prove that 2 2 2 2 2.a b c abc

Solution : We know 2a b c and squaring, we get 2 2 2 24 ( ) 2( )a b c a b c ab bc ca

2 2 2 2(2 )a b c ab bc ca

Adding 2abc if both sides, we get 2 2 2 2 2(2 )a b c abc ab bc ca abc

To prove 2 2 2 2,a b c abc it is enough to prove that

2(2 ) 2ab bc ca abc

or 2 ) 1abc ab bc ca

or 1 0ab bc ca abc 2 2a b c s 1s Example 48. For , 1,n N n show that

21 1 1 1... 1.1 2n n n n

Solution : We have

2 2

2 2 2 2 2

( ) terms ( ) terms

1 1 1 1 1 1 1 1 1... ...1 2

n n n n

n n n nn n n n n

2

2 21 1 1 1 1 ( ) 1 1... 1 1.1 2

n nn n n n n nn n

Example 49. Let a, b, c be real numbers with 0 , , 1a b c and 2.a b c Prove that

. . 8.1 1 1a b c

a b c

Solution : Here we use A.M. – G.M.

( ) ( ) ( )( )2a b c a b ca a b c a b c

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( ) ( ) ( )( )2b a c b a cb b a c b a c

( ) ( ) ( )( )2c a b c a bc c a b c a b

[( ) ( )][( ) ( )][( ) ( )]. . .8a b c a b c b a c b a c c a b c a ba b c

( )( )( )( )( )( )a b c a b c b a c b a c c a b c a b

(2 2 )(2 2 )(2 2 )c a b

8(1 )(1 )(1 ).c a b [ 2]a b c

. 8.1 1 1a b c

a b c