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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

` ge (a+b+c)`

B

`ge (1)/(2)(a+b+c)`

C

`(3)/(2) le (a+b+c)`

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
A
`(a)` `(a^(2)+b^(2))/(a+b)+(b^(2)+c^(2))/(b+c)+(a^(2)+c^(2))/(a+c)`
`=(a^(2))/(a+b)+(b^(2))/(b+c)+(c^(2))/(a+c)+(b^(2))/(a+b)+(c^(2))/(b+c)+(a^(2))/(a+c)`
`ge ((2a+2b+2c)^(2))/(4(a+b+c)) ge (a+b+c)`
`implies(a^(2))/(a+b)+(b^(2))/(b+c)+(c^(2))/(a+c)+(b^(2))/(a+b)+(c^(2))/(b+c)+(a^(2))/(a+c) ge a+b+c`
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