Let $N = {q^k}{n^2}$ be an odd perfect number given in Eulerian form.
That is, $q$ is prime with $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q, n) = 1$.
Since prime powers are not perfect and $\gcd(q, n) = 1$, then $q \neq n$ and $q^k \neq n$ both hold.
We want to prove the following proposition.
PROPOSITION 1: If $N = {q^k}{n^2}$ is an odd perfect number given in
Eulerian form with $q^k < n$, then the biconditional
$$q^k < n \Longleftrightarrow \sigma(q^k) < \sigma(n) \Longleftrightarrow \frac{\sigma(q^k)}{n} < \frac{\sigma(n)}{q^k}$$
holds.
PROOF:
Note that the inequation
$$\frac{\sigma(q^k)}{n} \neq \frac{\sigma(n)}{q^k}$$
is trivial, and follows from the fact that $\gcd(q, n) = 1$ and $1 < I(q^k)I(n) < 2$.
We consider three cases:
Case 1. $\frac{\sigma(q^k)}{n} + \frac{\sigma(n)}{q^k} < I(q^k) + I(n)$
Under this case, we have
$$\left(\sigma(q^k) - \sigma(n)\right)\left(q^k - n\right) < 0.$$
Consequently, we have the biconditional
$$q^k < n \Longleftrightarrow \sigma(n) < \sigma(q^k).$$
However, the conjunction
$$q^k < n < \sigma(n) < \sigma(q^k)$$
contradicts $I(q^k) < I(n)$ (which can be clearly seen when written in the following form):
$$q^k < n < \sigma(n) < \sigma(q^k)$$
contradicts $I(q^k) < I(n)$ (which can be clearly seen when written in the following form):
$$1 < \frac{\sigma(q^k)}{\sigma(n)} < \frac{q^k}{n} < 1.$$
On the other hand, the conjunction
$$n < q^k < \sigma(q^k) < \sigma(n)$$
is ruled out by the premise $q^k < n$.
Notice that
$$\sigma(q^k) = \sigma(n)$$
if and only if
$$\frac{\sigma(q^k)}{n} + \frac{\sigma(n)}{q^k} = I(q^k) + I(n).$$
Hence, if $q^k < n$, the inequality
$$I(q^k) + I(n) \leq \frac{\sigma(q^k)}{n} + \frac{\sigma(n)}{q^k}$$
holds.
Under this case, we have
Since $q^k \neq n$, this means that
On the other hand, the conjunction
$$n < q^k < \sigma(q^k) < \sigma(n)$$
is ruled out by the premise $q^k < n$.
Notice that
$$\sigma(q^k) = \sigma(n)$$
if and only if
$$\frac{\sigma(q^k)}{n} + \frac{\sigma(n)}{q^k} = I(q^k) + I(n).$$
Hence, if $q^k < n$, the inequality
$$I(q^k) + I(n) \leq \frac{\sigma(q^k)}{n} + \frac{\sigma(n)}{q^k}$$
holds.
Case 2. $\frac{\sigma(q^k)}{n} + \frac{\sigma(n)}{q^k} = I(q^k) + I(n)$
Under this case, we have
$$\left(\sigma(q^k) - \sigma(n)\right)\left(q^k - n\right) = 0.$$
Since $q^k \neq n$, this means that
$$\sigma(q^k) = \sigma(n)$$
which further implies that $n < q^k$ (since $I(q^k) < I(n)$), contradicting the premise $q^k < n$.
We are therefore left with:
We are therefore left with:
Case 3. $I(q^k) + I(n) < \frac{\sigma(q^k)}{n} + \frac{\sigma(n)}{q^k}$
Under this case, we have
$$\left(\sigma(q^k) - \sigma(n)\right)\left(n - q^k\right) < 0.$$
Consequently, we have the biconditional
$$q^k < n \Longleftrightarrow \sigma(q^k) < \sigma(n).$$
This then implies that we have the biconditionals
$$q^k < n \Longleftrightarrow \sigma(q^k) < \sigma(n) \Longleftrightarrow \frac{\sigma(q^k)}{n} < \frac{\sigma(n)}{q^k}$$
under the assumption $q^k < n$.
This then implies that we have the biconditionals
$$q^k < n \Longleftrightarrow \sigma(q^k) < \sigma(n) \Longleftrightarrow \frac{\sigma(q^k)}{n} < \frac{\sigma(n)}{q^k}$$
under the assumption $q^k < n$.
QED.
