The sum of submultiplcative function $\omega$ and ternary divisor function $\tau_3$

Question

Definition 1. An arithmetic function $\omega: \mathbb{N}\to \mathbb{R}_{\geq 0}$ is sub-multiplicative if $\omega(ab)\leq \omega(a)\omega(b)$ for all $a,b\in \mathbb{N}$ and whenever $d\mid n$ we have $\omega (d)\leq \omega(n)$.

Also $\tau_3(n)$ is the ternary divisor function $\sum \limits_{abc=n}1$.

I was wondering how to show that $$\sum \limits_{k'\ell'e=q}\omega(k'e)\tau_3(k'e)\leq \omega(q)\tau_3(q)^2.$$

This is what I have so far. One can see that $$\sum \limits_{k'\ell'e=q}\omega(k'e)\tau_3(k'e)=\sum\limits_{xy=q}\omega(x)\tau_3(x).$$

For each $x$ such that $xy=q$ we have $\omega(x)\leq \omega(q)$ since $x\mid q$. Hence $$\sum\limits_{xy=q}\omega(x)\tau_3(x)\leq \omega(q)\sum \limits_{xy=q}\tau_3(x).$$

How to show that the last sum is $\leq \tau_3(q)^2$?

Answer 1

All that's left is to observe that

• $\tau_3(x) \le \tau_3(q)$ for all $x\mid q$, and
• the number of summands is $\tau_2(q) \le \tau_3(q)$.