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.
 A: 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 $\mathbb{N}$.
A: *

*First show for $u$, $v$ powers of the same prime $p$.


*Now consider arbitrary $u$, $v$. Then
$$u = \prod_p u_p,  \ \ v = \prod_p v_p$$
decompose each into a product of powers of distinct primes. Note that  we have  $(uv)_p = u_p v_p$
Therefore
$$\tau_3(uv) =(\tau_3 \textrm{ multiplicative}) \prod_p \tau_3 ((u v)_p) = \prod_p \tau_3 ( u_p v_p) \le \prod_p (\tau_3( u_p ) \cdot \tau_3 (v_p)) = \prod_p \tau_3(u_p) \cdot \prod_p \tau_3(v_p) = \tau_3(u) \cdot \tau_3(v)$$
The inequality $\tau_3 ( u_p v_p) \le \prod_p (\tau_3( u_p )$ is true since $u_p$, $v_p$ are both powers of $p$, and we showed 1.
Note: we also have the inequality $\tau_2 (uv) \le \tau_2(u) \tau_2(v)$. Probably it works for all $\tau_k$.
A: Lemma. Any two factorizations of an integer $n$ have a common refinement. In other words, if we have factorizations $n=x_1\cdots x_r$ and $n=y_1\cdots y_s$ there are $z_{ij}\,$ ($\small 1\le i\le r,1\le j\le s$) for which
$$ \begin{array}{c|ccc}
& x_1 & \cdots & x_r \\ \hline
y_1 & z_{11} & \cdots & z_{1r} \\
\vdots & \vdots & \ddots & \vdots \\
y_s & z_{s1} & \cdots & z_{sr}
\end{array} $$
(Interpret this as saying the product of the $z_{ij}$s along any column gives the $x_j$ above it or along any row gives the $y_i$ left of it.) I leave it as an exercise to verify the $2\times 2$ case, and the $s\times r$ case follows by induction.

Define $T_3(n)=\{(a,b,c)\mid abc=n\}$, so $\tau_3(n)=|T_3(n)|$.
There is a surjection $T_3(u)\times T_3(v)\to T_3(uv)$ given by
$$ \big(\,(a_1,b_1,c_1),\,(a_2,b_2,c_2)\,\big)\mapsto (a_1a_2,b_1b_2,c_1c_2). $$
We can see it is onto, by applying the lemma to the equations $uv=abc$.
This implies $|T_3(u)||T_3(v)|\ge|T_3(uv)|$, or $\tau_3(uv)\le\tau_3(u)\tau_3(v)$.
