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.
If `abc=1` , then the minimum value of
`(1)/(a^(3)(b+c))+(1)/(b^(3)(a+c))+(1)/(c^(3)(a+b))` is

Options

A. `3`
B. `3//2`
C. `6`
D. `9`

Answer

B

Explanation

`(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)`