Improving the bound $q < n\sqrt{3}$ for an odd perfect number $N = {q^k}{n^2}$ given in Eulerian form (Note:  This has been cross-posted to MO.)
Let $N = {q^k}{n^2}$ be an odd perfect number given in Eulerian form (i.e., $q$ is prime with $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q, n) = 1$).
Therefore, $q \neq n$.  It follows that either $q < n$ or $n < q$.
Dris [2012] proved that $n < q$ implies Sorli's conjecture that $k = 1$.  By the contrapositive, $k > 1$ implies that $q < n$.
Acquaah and Konyagin [2012] showed that all the prime factors $r$ of $N$ satisfy $r < (3N)^{1/3}$.  In particular, if $k = 1$, then
$$q < (3N)^{1/3} \Longrightarrow q^3 < 3N = 3qn^2 \Longrightarrow q < n\sqrt{3}.$$
Therefore, regardless of the status of Sorli's conjecture, we know that
$$q < n\sqrt{3}$$
must be true.
Let $\sigma(x)$ be the sum of the divisors of the positive integer $x$.  Let 
$$I(x) = \sigma(x)/x$$
be the abundancy index of $x$.
We claim that
$$\frac{\sigma(q)}{n} \neq \frac{\sigma(n)}{q}.$$
Suppose to the contrary that
$$\frac{\sigma(q)}{n} = \frac{\sigma(n)}{q}.$$
It follows that
$$q\sigma(q) = n\sigma(n).$$
Since $\gcd(q, n) = 1$, then we have
$$n \mid \sigma(q)$$
and
$$q \mid \sigma(n)$$
so that
$$\frac{\sigma(q)}{n}, \frac{\sigma(n)}{q} \in \mathbb{N}.$$
Since $1 < qn$ is a proper factor of the (odd) perfect number $N = {q^k}{n^2}$, then $qn$ is deficient, and we have
$$1 < I(qn) = \frac{\sigma(q)}{n}\cdot\frac{\sigma(n)}{q} < 2.$$
This is a contradiction.
Now, in an e-mail sent by Ochem to Dris in April of 2013, the following result was communicated:
If $N = {q^k}{n^2}$ is an odd perfect number given in Eulerian form, then
$$I(n) > \left(\frac{8}{5}\right)^{\frac{\ln(4/3)}{\ln(13/9}} \approx 1.44440557369828207.$$
Since $q < n\sqrt{3}$ (in general), then ( since $n > {10}^{375}$ follows from $q^k < n^2$ (Dris [2012]) and $N > {10}^{1500}$ (Ochem and Rao [2012]) )
$$\sigma(q) = q + 1 << n\sqrt{3}$$
$$\frac{\sigma(q)}{n} << \sqrt{3}$$
$$\frac{\sigma(n)}{q} > \left(\frac{8}{5}\right)^{\frac{\ln(4/3)}{\ln(13/9}}\frac{n}{q} > {\frac{\sqrt{3}}{3}}\cdot\left(\frac{8}{5}\right)^{\frac{\ln(4/3)}{\ln(13/9}} \approx 0.83392794679369899.$$
In particular, if
$$\frac{\sigma(q)}{n} < \frac{\sigma(n)}{q},$$
so that
$$\frac{\sigma(q)}{n} < \sqrt{2},$$
then we have
$$\frac{\sigma(n)}{q} > \sqrt{\frac{\sigma(q)}{n}\cdot\frac{\sigma(n)}{q}} = \sqrt{I(q)I(n)} > \sqrt{I(n)} > \sqrt{\left(\frac{8}{5}\right)^{\frac{\ln(4/3)}{\ln(13/9)}}}.$$ 
But
$$\sqrt{\left(\frac{8}{5}\right)^{\frac{\ln(4/3)}{\ln(13/9)}}} \approx 1.20183425383797497745284556 > I(5).$$
My question is this -- Does anybody here have any bright ideas on how to improve the following bounds:


$$\frac{\sigma(q)}{n} << \sqrt{3}$$
$$\frac{\sigma(n)}{q} > {\frac{\sqrt{3}}{3}}\cdot\left(\frac{8}{5}\right)^{\frac{\ln(4/3)}{\ln(13/9)}}$$


Basically, I'm trying to get an upper bound for $\sigma(q)/n$ that's smaller than the lower bound for $\sigma(n)/q$, in order to force $\sigma(q)/n < \sigma(n)/q$.  As you can see, it's still a long way to go.  (For more details on the motivation behind pursuing a proof for this inequality, the interested reader is hereby referred to this preprint.)
Thanks everyone, and my apologies for the somewhat very long post. =)

Update - July 19 2016 
Brown has announced a proof for the inequality $q < n$, and a partial proof that $q^k < n$ holds "in many cases", here.

 A: 
Update - July 19 2016
Brown has announced a proof for the inequality $q < n$, and a partial proof that $q^k < n$ holds "in many cases", here.

The following occurred to me just now.  This is not a complete answer to my question, but I merely wanted to collect my thoughts into a single location:
Since $\sigma(q^k) \equiv k + 1 \pmod 4$, $k \equiv 1 \pmod 4$ and $n$ is odd, then $\sigma(q^k) \neq n$.  We consider two cases:
$$\sigma(q^k) < n$$
and
$$n < \sigma(q^k).$$
The first implies $q^k < n$, which further implies $q < n$.
Under the second case, we consider two sub-cases:


Sub-Case (1) : $$\frac{\sigma(q^k)}{n} < \frac{\sigma(n)}{q^k}$$


In this subcase, $\sigma(q^k) < \sqrt{2}n$.  This implies that $q < \sqrt{2}n$.
Additionally, note that under this subcase:
$$\frac{\sigma(q)}{n} \leq \frac{\sigma(q^k)}{n} < \frac{\sigma(n)}{q^k} \leq \frac{\sigma(n)}{q}$$


Sub-Case (2) : $$\frac{\sigma(n)}{q^k} < \frac{\sigma(q^k)}{n}$$


This implies that $n < q^k$, which implies that the biconditional $k = 1 \Longleftrightarrow n < q$ is true.
Under Sub-Case (2):
(a) If $k > 1$, then $q < n$.
(b) If $k = 1$, then (by Acquaah and Konyagin's estimate) $q < \sqrt{3}n$.
This is my best shot at this, so far.
