Problem

Source: Ukraine TST 2013 p5

Tags: inequalities, algebra



For positive $x, y$, and $z$ that satisfy the condition $xyz = 1$, prove the inequality $$\sqrt[3]{\frac{x+y}{2z}}+\sqrt[3]{\frac{y+z}{2x}}+\sqrt[3]{\frac{z+x}{2y}}\le \frac{5(x+y+z)+9}{8}$$