Question

If a, b, c `epsilon` R+, then `(bc)/(b+c) + (ac)/(a+c) + (ab)/(a+b)` is always ? a.)`<= 1/2(a+b+c)` b.)`>=1/3[sqrt(abc)]` c.)`<= 1/3(a+b+c)` d.)`>=1/2[sqrt(abc)]`

If a, b, c `epsilon` R+, then `(bc)/(b+c) + (ac)/(a+c) + (ab)/(a+b)` is always ?
a.)`<= 1/2(a+b+c)` b.)`>=1/3[sqrt(abc)]` c.)`<= 1/3(a+b+c)` d.)`>=1/2[sqrt(abc)]`

Answer 1

Let `y=(bc)/(b+c) + (ac)/(a+c) + (ab)/(a+b)`

Now

`1/2(a+b+c)-y`

`= 1/2(a+b+c)-(bc)/(b+c) - (ac)/(a+c) - (ab)/(a+b)`

`=1/4[2(a+b+c)-(4bc)/(b+c) - (4ac)/(a+c) - (4ab)/(a+b)]`

`=1/4[(b+c)-(4bc)/(b+c) +(c+a)- (4ac)/(a+c) +(a+b)- (4ab)/(a+b)]`

`=1/4[((b+c)^2-4bc)/(b+c) +((c+a)^2- 4ac)/(a+c) +((a+b)^2- 4ab)/(a+b)]`

`=1/4[(b-c)^2/(b+c) +(c-a)^2/(a+c) +(a-b)^2/(a+b)]>=0`

[As it is given that `a,b,c in R+`]

So

`1/2(a+b+c)-y>=0`

`=>y<=1/2(a+b+c)`

`=>(bc)/(b+c) + (ac)/(a+c) + (ab)/(a+b)<=1/2(a+b+c)`

Answer 2

See below.

Explanation:

`f(a,b,c) = (b c)/(b + c) + (a c)/(a + c) + (a b)/(a + b)`

is symmetric regarding its arguments so it have a maximum/minimum for `a=b=c=lambda` which is

`f(lambda,lambda,lambda) = 3/2lambda = 1/2(lambda+lambda+lambda)` so the answer is

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