2004 IMO Shortlist S07

If $a$ , $b$ , $c$ are three positive real numbers such that $ab+bc+ca = 1$ , prove that $$\sqrt[3]{ \frac{1}{a} + 6b} + \sqrt[3]{\frac{1}{b} + 6c} + \sqrt[3]{\frac{1}{c} + 6a } \leq \frac{1}{abc}.$$

如果 $a$ 、 $b$ 、 $c$ 是三个正实数，使得 $ab+bc+ca = 1$ ，则证明 $$\sqrt[3]{ \frac{1}{a} + 6b} + \sqrt[3]{\frac{1}{b} + 6c} + \sqrt[3]{\frac{1}{c} + 6a } \leq \frac{1}{abc}。$$