Hkg TST

If $n = p^k$ for some odd prime $p$ and positive integer $k$, then we have
$$\frac{\sigma(n)}{n} = \frac{1 + p + \cdots + p^k}{p^k} = 1 + \frac{1}{p} + \cdots + \frac{1}{p^k} < \frac{1}{1 - \frac{1}{p}} = \frac{p}{p-1} < 2.$$
If $n = p^k q^\ell$ for some odd primes $p, q$ and positive integers $k, \ell$, then we have
$$\frac{\sigma(n)}{n} = \frac{1 + p + \cdots + p^k}{p^k} \cdot \frac{1 + q + \cdots + q^\ell}{q^\ell} < \frac{p}{p-1} \cdot \frac{q}{q-1} \le \frac{3}{2} \cdot \frac{5}{4} < 2.$$
This shows $\sigma(n) < 2n$ if $n$ has at most 2 distinct prime factors. Therefore, in order that $n$ is a perfect number, it must have at least 3 distinct prime factors.