Problem
NT-B2-M10-P017 Sum of Squares of Divisors
#17
★★★★★ Level 5 of 5
Prove that for every \(n\ge 1\), \(\sigma_2(n)\ge n\tau(n)\), where \(\sigma_2(n)=\sum_{d\mid n}d^2\).
1001 Problems in Classical Number Theory (method inspiration) · Problem 489
Pair divisors \(d\) and \(\frac{n}{d}\), then use \(u^2+v^2\ge 2uv\).
For each pair of distinct divisors \(d\) and \(\frac{n}{d}\), we have \(d^2+\left(\frac{n}{d}\right)^2\ge 2n\). The pair contains two divisors, so its contribution is at least \(n\) per divisor. If \(n\) is a square, the central divisor \(\sqrt{n}\) contributes \(n\), exactly \(n\) for one divisor. Summing over all divisors gives \(\sigma_2(n)\ge n\tau(n)\).
A good blend of divisor functions and an inequality.