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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.

          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.

Lemma 5. If n | N, then
(σ(n))/(n)≤(σ(N))/(N),
where equality holds only if n = N.
Proof. Note that if d | N then kd = N for some k, so k = (N/d) | N. This argument
works just as well in reverse, so we have that d | N if and only if (N/d) | N, which
implies that
σ(N) = ∑d | N d = ∑d | N(N)/(d) = N ∑d | N(1)/(d)
If n is a proper divisor of N, we have
(σ(n))/(n) < ∑p | n(1)/(p) < ∑p | N(1)/(p) = (σ(N))/(N)
Otherwise, equality holds.

Added by Angela M.

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