Goutham wrote: Let $\tau(n)$ be the number of positive divisors of a natural number $n$, and $\sigma(n)$ be their sum. Find the largest real number $\alpha$ such that \[\frac{\sigma(n)}{\tau(n)}\ge\alpha(n)\] for all $n \ge 1$. Do I understand well? $\sigma(n)\ge\tau(n)$ with equality if $n=1?$ Then $\alpha = 1.$ I don't think this can be meant?

Anyway, it should be just $\alpha$ instead of $\alpha(n)$.

Thanks. It is actually $\alpha\times \sqrt{n}$.

http://www.artofproblemsolving.com/blog/63508

For any pair of divisors $c$ and $\frac{n}{c}$ we have $c+\frac{n}{c}\ge 2\sqrt {n}$ So $\frac{\sigma (n)}{\tau (n)}=\frac{\sum_{c|n}c}{\sum_{c|n}1}=\frac{\sum_{c|n}c+\frac{n}{c}}{2\sum_{c|n}1}\ge \frac{\sum_{c|n}2\sqrt{n}}{2\sum_{c|n}1}=\sqrt{n}$ So $\alpha=1$

So we shall look when does the function $g(n)=\frac{\sigma(n)}{\tau(n)}$ decrease. So in a sense we look at the factorization of $n$. We examine each prime power that divides $n$. So now we say that $p^a \parallel n$ We look when does the following hold: $$a+1 \geqslant 1 + p + p^2 + ... + p^a$$But we see that this doesn't hold for any prime numbers, nor for any natural number. So we see that this inequality grows fastly. So we also see $g(n)$ is a increasing function $\forall n \in \mathbb{N}$. So the maximum of $\alpha$ is attained when $n=1$. $$g(1)=1 \geqslant \alpha \implies max\{\alpha\} = 1$$

I'm sorry to bump this old thread but the LHS is basically the arithmetic mean of the divisors so AM-GM gives $\frac{\sigma(n)}{\tau(n)}\ge \sqrt[\tau(n)]{d_1\times d_2 \times \cdots d_{\tau(n)}} $ Now if $d|n$ is a divisor$\Rightarrow \frac{n}{d}$ is also a divisor and their product is $n$ so the product of all $d_i$'s is simply $n^{\tau(n)/2}$ and thus the inequality $\frac{\sigma(n)}{\tau(n)}\ge \sqrt{n}$ and hence $\alpha=1$