Problem 1789

To prove the inequality
$$\sum_{m=1}^n 5^{\omega(m)} \le \sum_{k=1}^n \left\lfloor \frac{n}{k} \right\rfloor \tau(k)^2 \le \sum_{m=1}^n 5^{\Omega(m)},$$
we define the following functions:
$$\chi(n) = 3^{\omega(n)}, \quad \phi(n) = \sum_{d \mid n} \tau(d), \quad \psi(n) = 3^{\Omega(n)}.$$

We claim that:
$$\chi(n) \leq \phi(n) \leq \psi(n).$$

Proof:

Since all functions $\chi$, $\phi$, and $\psi$ are multiplicative, it suffices to check the inequality for prime powers $p^k$.

For a prime power $p^k$, we have:
$$\chi(p^k) = 3^1 = 3, \quad \psi(p^k) = 3^k, \quad \phi(p^k) = \sum_{d \mid p^k} \tau(d) = \sum_{n=0}^k n = \frac{k(k+1)}{2}.$$

We need to prove:
$$3 \leq \frac{k(k+1)}{2} \leq 3^k \quad \text{for} \quad k \geq 1.$$

The left inequality is obvious for $k \geq 1$. To show the right inequality, we use induction.

Base Case:
For $k = 1$, we have:
$$\frac{1(1+1)}{2} = 1 \leq 3.$$

Induction Hypothesis:
Assume the inequality holds for some $k \geq 1$:
$$\frac{k(k+1)}{2} \leq 3^k.$$

Induction Step:
We need to show it for $k+1$:
$$\frac{(k+1)(k+2)}{2} \leq 3^{k+1}.$$

Using the induction hypothesis:
$$\frac{(k+1)(k+2)}{2} = \frac{k(k+1) + 2(k+1)}{2} = \frac{k(k+1)}{2} + (k+1) \leq 3^k + (k+1).$$

Since $k+1 \leq 2 \cdot 3^k$ for $k \geq 1$, we have:
$$3^k + (k+1) \leq 3^k + 2 \cdot 3^k = 3 \cdot 3^k = 3^{k+1}.$$

Thus, the induction step is complete, and we have:
$$3 \leq \frac{k(k+1)}{2} \leq 3^k.$$

Therefore, we have shown that:
$$\chi(n) \leq \phi(n) \leq \psi(n).$$

Finally, we note that the left-hand side of the original inequality is:
$$\sum_{m=1}^n \chi(m),$$
and the right-hand side is:
$$\sum_{m=1}^n \psi(m).$$

Using the well-known identity:
$$\left\lfloor \frac{n}{k} \right\rfloor = \sum_{d \mid k \leq n} 1,$$
we complete the proof.

The answer is: \boxed{\sum_{m=1}^n 5^{\omega(m)} \le \sum_{k=1}^n \left\lfloor \frac{n}{k} \right\rfloor \tau(k)^2 \le \sum_{m=1}^n 5^{\Omega(m)}}.