Prove following inequality Prove that $(\frac{2a}{b+c})^\frac{2}{3}+(\frac{2b}{a+c})^\frac{2}{3}+(\frac{2c}{a+b})^\frac{2}{3} ≥ 3$
What I tried was to use AM-GM for the left side of this inequality, what I got was
$3(\frac{8abc}{(a+b)(b+c)(c+a)})^\frac{2}{9}≥ 3$ and $(\frac{8abc}{(a+b)(b+c)(c+a)})^\frac{2}{9}≥ 1$ or just $\frac{8abc}{(a+b)(b+c)(c+a)}≥ 1$, but this isn't true.

Hint:
WLOG you may set $a+b+c=3$ and show instead that for $x\in[0,3]$,
$$\left(\frac{2x}{3-x}\right)^{2/3}\geqslant x$$
But this is equivalent to the obvious $x^2(4-x)(x-1)^2\geqslant0$. 
