parmenides51 wrote: For positive $x, y$, and $z$ that satisfy the condition $xyz = 1$, prove the inequality $$\sqrt[3]{\frac{x+y}{2z}}+\sqrt[3]{\frac{y+z}{2x}}+\sqrt[3]{\frac{z+x}{2y}}\le \frac{5(x+y+z)+8}{9}$$ Put $(1,1,1)$

oops, i mistyped 9 and 8, I am changing them

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

parmenides51 wrote: For positive $x, y$, and $z$ that satisfy the condition $xyz = 1$, prove the inequality $$\sqrt[3]{\frac{x+y}{2z}}+\sqrt[3]{\frac{y+z}{2x}}+\sqrt[3]{\frac{z+x}{2y}}\le \frac{5(x+y+z)+9}{8}$$ Use $$\sqrt[3]{\frac{x^3+y^3}{2}}\leq\frac{x^2+y^2}{x+y}.$$