Abundancy Index of Product is greater than Abundancy Index of Proper Factors

Theorem
Let $n \in \Z_{>0}$ be a composite number such that $n = r s$, where $r, s \in \Z_{>1}$.

Then:
 * $\dfrac {\map {\sigma_1} n} n > \dfrac {\map {\sigma_1} r} r$

and consequently also:
 * $\dfrac {\map {\sigma_1} n} n > \dfrac {\map {\sigma_1} s} s$

where $\sigma_1$ denotes the divisor sum function.

That is, the abundancy index of a composite number is strictly greater than the abundancy index of its divisors.

Proof
Consider the divisors of $r$.

Let $d \divides r$, where $\divides$ indicates divisibility.

We have that:
 * $d \divides n$

and also that:
 * $d s \divides n$

Thus:

Similarly for $s$.