# 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.

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

• Creative AM-GM! +1 – Macavity Oct 29 '18 at 12:04

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

• Perhaps a stupid question to ask, but what if x is for example 3.5? Then your inequality is not quite correct? – Severus156 Oct 28 '18 at 21:29
• @Severus156 $x\in [0,3]$, as once you’ve scaled to $a+b+c=3$, none among $a,b,c$ can exceed $3$… – Macavity Oct 29 '18 at 1:49
• Nice solution! +1. – Michael Rozenberg Oct 29 '18 at 2:23
• Oh, thank you for explanation @Macavity, nice solution indeed:) – Severus156 Oct 29 '18 at 7:39