Problem
ALG-B2-M09-P009 Cubic sum
#9
★★★☆☆ Level 3 of 5
Let \(a,b,c\ge0\), \(a+b+c=S\). Prove \[a^3+b^3+c^3\ge\frac{S^3}{9}.\]
Hint. Use Jensen for \(x^3\) on \(x\ge0\).
The function \(x^3\) is convex on \(x\ge0\). By Jensen, \(\frac{\sum a^3}{3}\ge\left(\frac{S}{3}\right)^3\). Therefore \(\sum a^3\ge S^3/9\). Equality holds at \(a=b=c=S/3\).
Shows how fixed sum works with powers.