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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.

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

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    $\begingroup$ Creative AM-GM! +1 $\endgroup$ – Macavity Oct 29 '18 at 12:04
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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 $x^2(4-x)(x-1)^2\geqslant0$.

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

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