In triangle ABC, prove that `sin""(A)/(2)sin""(B)/(2)sin""(C)/(2)le(1)/(8)` and hence, prove that `co sec ""(A)/(2)+co sec""(B)/(2)+co sec""(C)/(2)ge6`.

Explanation

We know that is triangle ABC,
`cos A+cos B+cos Cle3//2`
Now, `cosA+cosB+cosC=1+4sin""(A)/(2)sin""(B)/(2)sin""(C)/(2)`.
`therefore 1+4sin""(4)/(2)sin""(A)/(2)sin""(B)/(2)sin""(C)/(2)le""(3)/(2)`
So, `sin""(A)/(2)(B)/(2)sin""(C)/(2)le(1)/(8)`
Using `AMgeGM`, we get
`(co sec((A)/(2))+co sec((B)/(2))+co sec((C)/(2)))/(3)ge(co sec ((A)/(2))co sec ((B)/(2))co sec((C)/(2)))^(1//3)`
`rArr co sec((A)/(2))+co sec((B)/(2))+co sec ((C)/(2))ge3((1)/(sin((A)/(2))sin((B)/(2))sin((C)/(2))))^(1//3)`
`rArr co sec((A)/(2))+co sec((B)/(2))+co sec((C)/(2))ge3(8)^(1//3)`
`rArr co sec((A)/(2))+co sec((B)/(2))+co sec((C)/(2))ge6`.