Problem
ALG-B2-M04-P007 Cubes and squares
#7
★★★☆☆ Level 3 of 5
Prove for \(x,y,z\ge0\): \[3(x^3+y^3+z^3)\ge(x+y+z)(x^2+y^2+z^2).\]
Hint. Order \(x,y,z\). Then \(x\) and \(x^2\) are ordered in the same way.
After renaming, assume \(x\le y\le z\). Then \(x^2\le y^2\le z^2\). By Chebyshev, \[\frac{x^3+y^3+z^3}{3}\ge\frac{x+y+z}{3}\cdot\frac{x^2+y^2+z^2}{3}.\] Multiplying by \(9\) gives the result.
Stress that renaming is allowed because the inequality is symmetric.