TITLE: Prove following inequality
QUESTION [6 upvotes]: 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.

REPLY [4 votes]: By AM-GM
$$\sum_{cyc}\left(\frac{2a}{b+c}\right)^{\frac{2}{3}}=\sum_{cyc}\frac{1}{\sqrt[3]{\left(\frac{b+c}{2a}\right)^2\cdot1}}\geq\sum_{cyc}\frac{1}{\frac{\frac{b+c}{2a}+\frac{b+c}{2a}+1}{3}}=\sum_{cyc}\frac{3a}{a+b+c}=3.$$