Question 20
For each positive integer $N$, let $\tau(N)$ be the number of positive factors of $N$; $\omega(N)$ be the number of distinct prime factors of $N$; $\Omega(N)$ be the number of prime factors (counts multiplicities) of $N$. Prove: for each positive integer $n$,
$$
\sum_{m=1}^{n} 5^{\omega(m)} \le \sum_{k=1}^{n} \lfloor \frac{n}{k} \rfloor \tau(k)^2 \le \sum_{m=1}^{n} 5^{\Omega(m)}.
$$
Here, $\lfloor x \rfloor$ is the largest integer not exceeding $x$.
$$
\sum_{m=1}^{n} 5^{\omega(m)} \le \sum_{k=1}^{n} \lfloor \frac{n}{k} \rfloor \tau(k)^2 \le \sum_{m=1}^{n} 5^{\Omega(m)}.
$$
Here, $\lfloor x \rfloor$ is the largest integer not exceeding $x$.