MathDB

\sqrt[3]{(x+y)/2z}+\sqrt[3]{(y+z)/2y}+\sqrt[3]{(z+x)/2y} <= [5(x+y+z)+9]/8

Source: Ukraine TST 2013 p5

April 28, 2020

inequalitiesalgebra

Problem Statement

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}