What is the solution to $\frac1{a^2 +2} + \frac1{b^2 +2} + \frac1{c^2 +2} \le \frac{\sqrt2}{2}\frac{\sqrt a+\sqrt b+\sqrt c}{\sqrt{abc}}$ one of my friends asked me if I could solve him a mathematics problem.
It looks like this:
$$\frac1{a^2 +2} + \frac1{b^2 +2} + \frac1{c^2 +2} \le \frac{\sqrt2}{2}\frac{\sqrt a+\sqrt b+\sqrt c}{\sqrt{abc}}$$
As I think it looks like an inequality between means, hope it helps.
And sorry for my bad english by the way. :)

@michael Rozenberg: Let's bury this one for good by presenting a point with all positive $a,b,c$ such that the inequality is violated.
Take
$$
a = 25\\
b=25\\
c= \frac{59}{81} \\
\frac{1}{a^2+2}+\frac{1}{b^2+2}+\frac{1}{c^2+2} = \frac{4146953}{10410081}
> 0.3983 \\
\frac{\sqrt{2}}{2}\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{\sqrt{abc}}=
\frac{59+90\sqrt{59}}{2950}\sqrt{2} < 0.3597
\\\mbox{LHS } > 0.39 > 0.36 > \mbox{ RHS}
$$
which violates the inequality.
