Problem
ALG-B2-M04-P011 Cubic lower bound
#11
★★★★☆ Level 4 of 5
Prove for \(a,b,c\ge0\): \[a^3+b^3+c^3\ge\frac{(a+b+c)^3}{9}.\]
Hint. First apply Chebyshev to \(a\) and \(a^2\), then use \(a^2+b^2+c^2\ge\frac{(a+b+c)^2}{3}\).
From the cube inequality, \[3\sum a^3\ge\left(\sum a\right)\left(\sum a^2\right).\] Also, \(\sum a^2\ge\frac{(\sum a)^2}{3}\). Therefore \[3\sum a^3\ge \sum a\cdot\frac{(\sum a)^2}{3}=\frac{(\sum a)^3}{3},\] so \(\sum a^3\ge\frac{(\sum a)^3}{9}\).
This already has two steps: Chebyshev plus a basic quadratic estimate.