Q:

Prove $d\left( ab\right) \geq d\left( a\right) +d\left( b\right) -7$

How do I prove that $d\left( ab\right) \geq d\left( a\right) +d\left( b\right) -7$?
Here, $d(n)$ is the number of divisors of $n$.

A:

We can prove something stronger. In fact: $$d(n)\geq d(a)+d(b)-1.$$
To do this, let $p_1$, $p_2$, ..., $p_n$ be the prime factors of either $a$ or $b$. We can write $$a=p_1^{\alpha_1}p_2^{\alpha_2}\ldots p_n^{\alpha_n},$$ $$b=p_1^{\beta_1}p_2^{\beta_2}\ldots p_n^{\beta_n},$$ so that $\alpha_i\geq0$, $\beta_i\geq0$, for all $i$. By using the formula for the number of divisors of a number, what we want to prove is equivalent to $$\prod_{i=1}^n \left(\alpha_i+\beta_i+1\right)\geq \prod_{i=1}^n \left(\alpha_i+1\right)+\prod_{i=1}^n \left(\beta_i+1\right)-1.$$ However, when we expand this out, it's easy to see that every summand on the RHS will appear too on the LHS, except for the $1$'s, which cancel out anyway. Since every summand is non-negative, this is enough to prove our inequality.

