Let $a,b,c > 0$ and $ab+bc+ca = 1$. Prove the inequality $3\sqrt[3]{\frac{1}{abc} +6(a+b+c) }\le \frac{\sqrt[3]3}{abc}$
Source: Slovenia TST 2005 p6
Tags: inequalities, algebra
Let $a,b,c > 0$ and $ab+bc+ca = 1$. Prove the inequality $3\sqrt[3]{\frac{1}{abc} +6(a+b+c) }\le \frac{\sqrt[3]3}{abc}$
