# Another inequality involving sigma and tau

Let $n$ be a positive integer. By the weighted AM-GM inequality, one has

$\displaystyle\frac{\sum d\cdot n/d}{\sum n/d}\ge\left(\prod d^{n/d}\right)^\frac{1}{\sum n/d}$,

where all sums and products are taken over the positive divisors $d$ of $n$. This means

$\displaystyle\boxed{\frac{n\tau(n)}{\sigma(n)}\ge\left(\prod_{d\mid n}d^{1/d}\right)^{n/\sigma(n)}}$.

Considering the analytic behaviour of the function $f(x)=x^{1/x}$ one deduces

$\displaystyle\prod_{d\mid n}d^{1/d}\ge\prod_{1\neq d\mid n}n^{1/n}=n^{(\tau(n)-1)/n}$

for any $n$, with equality iff $n$ is prime or $1$ or $4$. (By convention we take an empty product to be $1$.) Combining this with the first inequality in this post we obtain

$\displaystyle n^{1/2}\ge\frac{n\tau(n)}{\sigma(n)}\ge n^\frac{\tau(n)-1}{\sigma(n)}$,

i.e.,

$\boxed{\displaystyle n^{1/2}\le\frac{\sigma(n)}{\tau(n)}\le n^{1-\frac{\tau(n)-1}{\sigma(n)}}}$.

This strengthens the first boxed inequality from this post.