Inequality $\tau_3(uv)\leq \tau_3(u)\tau_3(v)$ for ternary divisor function Consider ternary divisor function $\tau_3:\mathbb{N}\to \mathbb{R}_{\geq}0$ defined as $$\tau_3(n)=\#\{(a,b,c)\in \mathbb{N}^3:abc=n\}.$$
I know that $\tau_3$ is multiplicative function, i.e. $\tau_3(mn)=\tau_3(m)\tau_3(n)$ for $\gcd(m,n)=1$. Also one can show that $$\tau_3(p^{\alpha})=\frac{(\alpha+1)(\alpha+2)}{2}.$$
I was wondering how to show that $\tau_3(uv)\leq \tau_3(u)\tau_3(v)$ for any $u,v\in \mathbb{N}$?
Can anyone show the proof please.

Since $\tau_3$ is multiplicative
$$ \tau_3(n) = \prod_{p\in\mathcal{P}}\frac{(\nu_p(n)+2)(\nu_p(n)+1)}{2} $$
hence in order to show that $\tau_3(uv)\leq \tau_3(u)\tau_3(v)$ it is enough to show that
$$\frac{(\nu_p(u)+\nu_p(v)+2)(\nu_p(u)+\nu_p(v)+1)}{2}\leq \frac{(\nu_p(u)+2)(\nu_p(u)+1)}{2}\cdot\frac{(\nu_p(v)+2)(\nu_p(v)+1)}{2}$$
or
$$ 2(A+B+2)(A+B+1) \leq (A+2)(A+1)(B+2)(B+1) $$
or
$$ AB(AB+3A+3B+5)\geq 0 $$
which is fairly trivial over $\mathbb{N}$.
