# An inequality involving sigma and tau

Let $\tau(n)$ and $\sigma(n)$ respectively be the number and the sum of the positive divisors of the positive integer $n$. We will prove the following inequality:

$\sigma(n)\ge \sqrt n\tau(n)$.

First proof. By the Cauchy-Schwarz inequality,

$\displaystyle\sigma(n)^2=\left(\sum_{d\mid n}d\right)\left(\sum_{d\mid n}\frac nd\right)\ge (\sqrt n\tau(n))^2=n\tau(n)^2$.

Second proof. By Chebyshev’s inequality,

$\displaystyle\sigma(n)^2=\left(\sum_{d\mid n}d\right)\left(\sum_{d\mid n}\frac nd\right)\ge \tau(n)\cdot n\tau(n)$.

Third proof. By the AM-GM inequality,

$\displaystyle 2\sigma(n)=\sum_{d\mid n}\left(d+\frac nd\right)\ge 2\sqrt n\tau(n)$.

Fourth proof. Again by the AM-GM inequality,

$\displaystyle n^{\tau(n)}=\prod_{d\mid n}d\prod_{d\mid n}n/d=\left(\prod_{d\mid n}d\right)^2\Rightarrow n^{\tau(n)/2}=\prod_{d\mid n}d\le\left(\frac{\sigma(n)}{\tau(n)}\right)^{\tau(n)}$.

Fifth proof. If $n=p_1^{a_1}\cdots p_k^{a_k}$ factored into primes, then AM-GM gives

$\displaystyle\sigma(n)=\prod_{j=1}^k(p_j^{a_j}+\cdots+p_j+1)\ge\prod_{j=1}^k(a_j+1)p_j^{a_j/2}=\sqrt{n}\tau(n)$.

We also trivially have $\sigma(n)\le n\tau(n)$. Hence for all $n$,

$\displaystyle \boxed{\sqrt n\le\frac{\sigma(n)}{\tau(n)}\le n}$.

In general, if $f:\mathbb N\to\mathbb N$ is completely multiplicative, and

$F(n):=\displaystyle\sum_{d\mid n}f(d)$,

then the first four proofs can be generalised to deduce that

$(*)\qquad\qquad\qquad\qquad\quad\ F(n)\ge \tau(n)\sqrt{f(n)}$.

For example, if $\sigma_k(n)$ is the sum of the $k$-th powers of the positive divisors of $n$, then this shows that

$\displaystyle \boxed{n^{k/2}\le\frac{\sigma_k(n)}{\tau(n)}\le n^k}$

for each $k\ge 0$. This, combined with the result from the previous post, gives

$\displaystyle\lim_{n\to\infty}\frac{\sigma_k(n)}{n^{k+1}}=0$,

i.e., $\sigma_k(n)=o(n^{k+1})$ as $n\to\infty$, for each $k\ge 0$.

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