Lemma 5. If $n \mid N$, then
$\frac{\sigma(n)}{n} \le \frac{\sigma(N)}{N}$,
where equality holds only if $n = N$.
Proof. Note that if $d \mid N$ then $kd = N$ for some $k$, so $k = (N/d) \mid N$. This argument
works just as well in reverse, so we have that $d \mid N$ if and only if $(N/d) \mid N$, which
implies that
$\sigma(N) = \sum_{d \mid N} d = \sum_{d \mid N} \frac{N}{d} = N \sum_{d \mid N} \frac{1}{d}$
If $n$ is a proper divisor of $N$, we have
$\frac{\sigma(n)}{n} < \sum_{p \mid n} \frac{1}{p} < \sum_{p \mid N} \frac{1}{p} = \frac{\sigma(N)}{N}$
Otherwise, equality holds.
