ALG-B2-M04-P022 Fifth power via the sum

Hint. Use a chain involving \(\sum a^5\), \(\sum a^3\), \(\sum a^2\), and \(\sum a\).

By Chebyshev for \(a^2\) and \(a^3\), \[3\sum a^5\ge(\sum a^2)(\sum a^3).\] Also \(3\sum a^3\ge(\sum a)(\sum a^2)\), and \(\sum a^2\ge\frac{(\sum a)^2}{3}\). Hence \(\sum a^3\ge\frac{(\sum a)^3}{9}\). Substituting this and using \(\sum a^2\ge\frac{(\sum a)^2}{3}\) again, we get \[3\sum a^5\ge\frac{(\sum a)^2}{3}\cdot\frac{(\sum a)^3}{9}=\frac{(\sum a)^5}{27}.\] Therefore \(\sum a^5\ge\frac{(\sum a)^5}{81}\).

A multi-step problem: the student must combine two Chebyshev applications with a quadratic estimate.