# An Inequality from RMM with a Generic 5

### Problem

### Proof

By *Hölder's inequality*,

$\begin{align} 2\sqrt[3]{abc}+5 &= \sqrt[3]{8abc}+\sqrt[3]{125}\\ &\le \sqrt[3]{2a+5}\sqrt[3]{2b+5}\sqrt[3]{2c+5}\\ &=\sqrt[3]{(2a+5)(2b+5)(2c+5)}. \end{align}$

On the other hand, $\displaystyle 2\sqrt[3]{abc}\ge\frac{6abc}{ab+bc+ca}.\;$ Indeed, by the AM-GM inequality, the latter is equivalent to

$\begin{align} 2\sqrt[3]{abc}(ab+bc+ca) &\ge 2\sqrt[3]{abc}\left(3\sqrt[3]{(abc)^2}\right)\\ &= 6abc. \end{align}$

### Acknowledgment

Dan Sitaru has kindly posted the above problem from the Romanian Mathematical Magazine, with two practically identical proofs - one by Kevin Soto Palacios (Peru), the other by Pham Quy (Vietnam), at the CutTheKnotMath facebook page.

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