Prove following inequality

By AM-GM $$\sum_{cyc}\left(\frac{2a}{b+c}\right)^{\frac{2}{3}}=\sum_{cyc}\frac{1}{\sqrt[3]{\left(\frac{b+c}{2a}\right)^2\cdot1}}\geq\sum_{cyc}\frac{1}{\frac{\frac{b+c}{2a}+\frac{b+c}{2a}+1}{3}}=\sum_{cyc}\frac{3a}{a+b+c}=3.$$

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 $x^2(4-x)(x-1)^2\geqslant0$.