If `a ,b ,c` are the sides of a triangle, then the minimum value of `a/(b+c-a)+b/(c+a-b)+c/(a+b-c)` is equal to `3` `6` `9` `12`

Options

A. 3
B. 6
C. 9
D. 12

Answer

A

Explanation

`2E = (2a)/(b + c - a) + (2b)/(c + a - b) + (2c)/(a + b - c)`
`= (2a)/(b + c + a) + 1 + (2b)/(c + a + b) + 1 + (2c)/(a + b -c) + 1 - 3`
`= (a + b + c) ((1)/(b + c -a) + (1)/(c + a - b) + (1)/(a + b - c)) - 3`
Using A.M `ge` H.M we have
`((1)/(b + c - a) + (1)/(c + a - b) + (1)/(a + b -c))/(3) ge (3)/(a + b + c)`
Or `(a + b+ c) ((1)/(b + c -a)+(1)/(c+a+b) + (1)/(a + b+ c)) ge 9`
or `(a + b+ c) ((1)/(b + c -a) + (1)/(c + a - b) + (1)/(a + b - c)) - 3 ge 6`
`implies E ge 3`