The number of divisors of any positive number $n$ is $\le 2\sqrt{n}$

Question

How to prove that the number of divisors of any positive number $n$ is $\le 2\sqrt{n}$?

I have started something like below:

$$ n^{\tau(n)/2} = \prod_{d|n} d$$

But not getting ideas on how to take this further and conclude. Any help is appreciated. Thanks!

Answer 1

There are at most $\sqrt n$ divisors $d\mid n$ with $d\le \sqrt n$, and there are at most $\sqrt n$ divisors with $d>\sqrt n$ as each of them corresponds uniquely to a divisor $\frac nd<\sqrt n$.