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If a,b,x,y are real number and x,y gt 0,...

If `a,b,x,y` are real number and `x,y gt 0`, then `(a^(2))/(x)+(b^(2))/(y) ge ((a+b)^(2))/(x+y)` so on solving it we have `(ay-bx)^(2) ge 0`.
Similarly, we can extend the inequality to three pairs of numbers, i.e,
`(a^(2))/(x)+(b^(2))/(y)+(c^(2))/(z) ge ((a+b+c)^(2))/(x+y+z)`
Now use this result to solve the following questions.
The value of `(a^(2)+b^(2))/(a+b)+(b^(2)+c^(2))/(b+c)+(a^(2)+c^(2))/(a+c)` is

A

`3`

B

`3//2`

C

`6`

D

`9`

Text Solution

Verified by Experts

The correct Answer is:
B
`(b)` `(1)/(a^(3)(b+c))+(1)/(b^(3)(a+c))+(1)/(c^(3)(a+b))`
`=(1//a^(2))/(ab+ac)+(1//b^(2))/(ab+bc)+(1//c^(2))/(ac+bc)`
`ge (((1)/(a)+(1)/(b)+(1)/(c ))^(2))/(2(ab+bc+ac))` ……….`(i)`
Now `((1)/(a)+(1)/(b)+(1)/(c ))^(2)=((ab+bc+ac)^(2))/((abc)^(2))`
`=(ab+bc+ac)^(2)` (as `abc=1`)
`:.` From `(i)`, `(1)/(a^(3)(b+c))+(1)/(b^(3)(a+c))+(1)/(c^(3)(a+b))`
` ge((ab+bc+ac))/(2)`
`ge(3*3sqrt((abc)^(2)))/(2)` (using `A.M. ge G.M.`)
`ge (3)/(2)`
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