Problem

Source: 2021ChinaTST test3 day2 P1

Tags: number theory function, number theory, Innequality, floor function, inequalities



Proof that $$ \sum_{m=1}^n5^{\omega (m)} \le \sum_{k=1}^n\lfloor \frac{n}{k} \rfloor \tau (k)^2 \le \sum_{m=1}^n5^{\Omega (m)} .$$