[ Number of divisors and sum of divisors of a number ]

Let $(m, n)>1$. Which is greater, $\tau(m n)$ or $\tau(m) \tau(n)$? Investigate the same question for the function $\sigma(n)$.
From the multiplicativity of the functions $\tau(n)$ and $\sigma(n)$, it follows that to justify the answer, it is sufficient to check the inequalities for the case when the numbers $m$ and $n$ are powers of the same prime number. Let $m=p^{\alpha}, n=p^{\beta}$. Then (see problem 60537) $\tau(m) \tau(n)=(\alpha+1)(\beta+1)>\alpha+\beta+1=\tau(m n)$ $\sigma(m) \sigma(n)>\sigma(m n) \Leftrightarrow\left(p^{\alpha+1}-1\right)\left(p^{\beta+1}-1\right)>\left(p^{\alpha+\beta+1}-1\right)(p-1) \Leftrightarrow p^{\alpha+\beta+1}+p>p^{\alpha+1}+p^{\beta+1} \Leftrightarrow p\left(p^{\alpha}-1\right)\left(p^{\beta}-1\right)>$ 0, and the last inequality is obvious.

## Answer

$\tau(m n)<\tau(m) \tau(n) ; \sigma(m n)<\sigma(m) \sigma(n)$.