On odd perfect numbers $N$ given in the Eulerian form $N = {q^k}{n^2}$ Note: This question was cross-posted from MO.
Preamble:  I apologize in advance if this particular MSE post would appear to be a bit of a polymath approach, I just had to put down all the details to present my argument for this particular math problem.
A positive integer $N$ is said to be perfect if $\sigma(N)=2N$, where $\sigma(x)$ is the sum of the divisors of $x$.
An odd perfect number $N$ is said to be given in Eulerian form if $N = {q^k}{n^2}$, where $q$ is prime, $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
Since $\gcd(q,n)=1$ and prime powers are deficient, we have $q \neq n$ and $q^k \neq n$.
In an earlier version of the paper The Abundancy Index of Divisors of Odd Perfect Numbers (see here), it was conjectured that the biconditional
$$q^k < n \Longleftrightarrow \sigma(q^k) < \sigma(n) \Longleftrightarrow \frac{\sigma(q^k)}{n} < \frac{\sigma(n)}{q^k}$$
is true.  (Note that the proof of the inequation
$$\frac{\sigma(q^k)}{n} \neq \frac{\sigma(n)}{q^k}$$
is trivial.)
Recently, an attempt to prove the said biconditional appears to have been completed in this preprint.
I present here the highlights of the said proof:


One direction of the biconditional is trivial:
$$q^k < n \Longrightarrow \frac{\sigma(q^k)}{n} < \frac{\sigma(n)}{q^k}.$$
This is proved by noting that $I(q^k) < \sqrt[3]{2} < I(n)$ (where $I(x) = \sigma(x)/x$ is the abundancy index of $x$), from which the following chain of implications follow:
$$q^k < n \Longrightarrow \sigma(q^k) < \sigma(n)$$
    $$q^k < n \Longrightarrow \frac{1}{n} < \frac{1}{q^k}$$
    $$\{\sigma(q^k) < \sigma(n)\} \land \{\frac{1}{n} < \frac{1}{q^k}\} \Longrightarrow \frac{\sigma(q^k)}{n} < \frac{\sigma(n)}{q^k}$$
Therefore:
$$q^k < n \Longrightarrow \sigma(q^k) < \sigma(n) \Longrightarrow \frac{\sigma(q^k)}{n} < \frac{\sigma(n)}{q^k}.$$


For the other direction:


The implication
$$\frac{\sigma(q^k)}{n} < \frac{\sigma(n)}{q^k} \Longrightarrow \sigma(q^k) < \sigma(n)$$
    can be proved, again by observing that $I(q^k) < \sqrt[3]{2} < I(n)$.
Now, to prove the last implication:
$$\sigma(q^k) < \sigma(n) \Longrightarrow q^k < n$$
we take an indirect approach.


First we show that:


If $N = {q^k}{n^2}$ is an odd perfect number given in Eulerian form and
$$I(q^k) + I(n) < \frac{\sigma(q^k)}{n} + \frac{\sigma(n)}{q^k},$$
then $q^k < n \Longleftrightarrow \sigma(q^k) < \sigma(n)$.


Similarly, we can prove that:


If $N = {q^k}{n^2}$ is an odd perfect number given in Eulerian form and
$$\frac{\sigma(q^k)}{n} + \frac{\sigma(n)}{q^k} < I(q^k) + I(n),$$
then $q^k < n \Longleftrightarrow \sigma(n) < \sigma(q^k)$.


Observe that
$$I(q^k) + I(n) = \frac{\sigma(q^k)}{n} + \frac{\sigma(n)}{q^k}$$
if and only if
$$\sigma(q^k) = \sigma(n).$$
Also, observe that if we assume
$$\frac{\sigma(q^k)}{n} + \frac{\sigma(n)}{q^k} < I(q^k) + I(n)$$
then the biconditional
$$q^k < n \Longleftrightarrow \sigma(n) < \sigma(q^k)$$
will contradict $I(q^k) < \sqrt[3]{2} < I(n)$.


Therefore, the following inequality must be true:
$$I(q^k) + I(n) \leq \frac{\sigma(q^k)}{n} + \frac{\sigma(n)}{q^k}.$$


It suffices to consider the case when
$$I(q^k) + I(n) = \frac{\sigma(q^k)}{n} + \frac{\sigma(n)}{q^k}$$
which is true if and only if
$$\sigma(q^k) = \sigma(n).$$
This last equation, together with the inequality $I(q^k) < \sqrt[3]{2} < I(n)$, implies that
$$1 = \frac{\sigma(q^k)}{\sigma(n)} < \frac{q^k}{n}$$
from which it follows that $n < q^k$. Thus, $1/q^k < 1/n$, which then gives, together with the equation $\sigma(n) = \sigma(q^k)$, the inequality
$$\frac{\sigma(n)}{q^k} < \frac{\sigma(q^k)}{n}.$$
But this last inequality, together with $I(q^k) < \sqrt[3]{2} < I(n)$, is known to imply
$$n < q^k.$$


Consequently, if
$$I(q^k) + I(n) = \frac{\sigma(q^k)}{n} + \frac{\sigma(n)}{q^k}$$
then
$$n < q^k \Longleftrightarrow \frac{\sigma(n)}{q^k} < \frac{\sigma(q^k)}{n}$$ 


Now, here is the part where I am a bit unsure about logical solidity:
Lastly, note that since
$$\sigma(q^k) = \sigma(n)$$
implies $n < q^k$, the biconditional
$$q^k < n \Longleftrightarrow \sigma(q^k) < \sigma(n)$$
is vacuously true, under this case.


NOW HERE IS MY QUESTION (and this is also the main reason for this MSE post):
I have been told that there is a gap in the last part of this proof.  As I
    myself am having a hard time spotting where that particular error is, would
    somebody be kind enough as to help by skimming through this argument and then 
    (after glossing over the details) give it either a PASS or a FAIL?


 A: Update (January 30, 2015)
Let $N = {q^k}{n^2}$ be an odd perfect number given in Eulerian form.
While $q^k < n$ does imply $\sigma(q^k) < \sigma(n)$ (via $I(q^k) < I(n)$), $\sigma(q^k) < \sigma(n)$ by itself DOES NOT imply
$$\frac{\sigma(q^k)}{n} < \frac{\sigma(n)}{q^k}.$$
Thus, the implication
$$\sigma(q^k) < \sigma(n) \implies \frac{\sigma(q^k)}{n} < \frac{\sigma(n)}{q^k}$$
remains unproven.
Similarly, while
$$\frac{\sigma(q^k)}{n} < \frac{\sigma(n)}{q^k}$$
does imply $\sigma(q^k) < \sigma(n)$ (via $I(q^k) < I(n)$), $\sigma(q^k) < \sigma(n)$ by itself DOES NOT imply $q^k < n$.
Thus, the implication $\sigma(q^k) < \sigma(n) \implies q^k < n$ remains unproven.
Further Update (July 14, 2016)

Brown has released a preprint titled A Partial Proof of a Conjecture of Dris where he announced a proof for the inequality $q < n$, as well as a partial proof that $q^k < n$ holds "in many cases".
If the inequality $q^k < n$ is proved, it will follow that the biconditionals
$$q^k < n \iff \sigma(q^k) < \sigma(n) \iff \dfrac{\sigma(q^k)}{n} < \dfrac{\sigma(n)}{q^k}$$
are also true.

