If
![a, b, c](/media/m/9/e/9/9e9dfe78930065fbe5a777e9b07c27c4.png)
are three positive real numbers such that
![ab+bc+ca = 1](/media/m/3/1/5/315be27fca0bcae6050ae5a6015fe406.png)
, prove that
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If $a, b, c$ are three positive real numbers such that $ab+bc+ca = 1$, prove that $$\sqrt[3]{ \frac{1}{a} + 6b} + \sqrt[3]{\frac{1}{b} + 6c} + \sqrt[3]{\frac{1}{c} + 6a } \leq \frac{1}{abc}.$$