If `a`, `b`, `c` are the sides of triangle , then the least value of `(a)/(c+a-b)+(b)/(a+b-c)+(c )/(b+c-a)` is

Options

A. `1//3`
B. `1`
C. `3`
D. `6`

Answer

C

Explanation

`(b)` `c+a-b`, `b+c-a`,`a+b-c` are all positive
Using `A.M. ge G.M.`
`:.(a)/(c+a-b)+(b)/(a+b-c)+(c )/(b+c-a) ge [(abc)/((c+a-b)(a+b-c)(b+c-a))]^(1//3)`........`(i)`
Also , `a^(2) ge a^(2)-(b-c)^(2)`
`implies a^(2) ge (a+b-c)(a-b+c)`
Similarly `b^(2) ge (b+c-a)(b-c+a)`
`c^(2) ge (c+a-b)(c-a+b)`
`:.a^(2)b^(2)c^(2) ge (a+b-c)^(2)(b+c-a)^(2)(c+a-b)^(2)`
Thus `abc ge (a+b-c)(b+c-a)(c+a-b)`
`implies (abc)/((c+a-b)(a+b-c)(b+c-a)) ge 1`
Hence, from `(i) (abc)/((c+a-b)(a+b-c)(b+c-a)) ge 1`