inequality

[Steven G]

I thing my mathematics are enough for that inequality

Great, so do the problem.

 

[chrislav]

From the C-S-B inequality we have:

[MATH](a+b+b+c+c+a)[\frac{1}{b(a+b)}+\frac{1}{c(b+c)}+\frac{1}{a(c+a)}\geq[\frac{1}{\sqrt b}+\frac{1}{\sqrt c}+\frac{1}{\sqrt a}]^2[/MATH]
And the rest of the proof is easy