Improving the bound for $\sigma(q^k)/q^k$ where $q^k n^2$ is an odd perfect number given in Eulerian form Let $x$ be a positive integer.  (That is, let $x \in \mathbb{N}$.)
We denote the sum of divisors of $x$ as
$$\sigma(x) = \sum_{d \mid x}{d}.$$
We also denote the abundancy index of $x$ as $I(x)=\sigma(x)/x$.
If $N$ is odd and $\sigma(N)=2N$, then $N$ is called an odd perfect number.  Euler proved that an odd perfect number, if one exists, must have the so-called Eulerian form $N = q^k n^2$, i.e. $q$ is the special prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
Since $k \equiv 1 \pmod 4$, then we have
$$\frac{q+1}{q} = I(q) \leq I(q^k) = \frac{q^{k+1} - 1}{q^k (q - 1)} < \frac{q^{k+1}}{q^k (q - 1)} = \frac{q}{q - 1}.$$
Since $q$ is prime and satisfies $q \equiv 1 \pmod 4$, we have $q \geq 5$, from which we obtain
$$\frac{1}{q} \leq \frac{1}{5} \implies -\frac{1}{q} \geq - \frac{1}{5} \implies \frac{q-1}{q} = 1 - \frac{1}{q} \geq 1 - \frac{1}{5} = \frac{4}{5}.$$
We get that
$$I(q^k) < \frac{q}{q - 1} \leq \frac{5}{4}$$
from which we conclude that
$$I(q^k) < \frac{5}{4}.$$
We also obtain
$$I(q^k) < \frac{5}{4} < \sqrt{\frac{8}{5}} < \sqrt{I(n^2)} = \sqrt{\frac{2}{I(q^k)}},$$
from which we get
$$\bigg(I(q^k)\bigg)^2 < \frac{2}{I(q^k)} \implies \bigg(I(q^k)\bigg)^3 < 2 \implies I(q^k) < \sqrt[3]{2}.$$
Here is my question:

Is it possible to improve on the inequality
  $$I(q^k) < \sqrt{\frac{2}{I(q^k)}},$$
  to something like (say)
  $$I(q^k) < \sqrt[3]{\frac{2}{I(q^k)}}?$$

Note that $1/3 = 0.\overline{333}$.
MOTIVATION
Here is the reason why I think it might be possible to improve on the bound for $I(q^k)$.
Let $N = q^k n^2$ be an odd perfect number given in Eulerian form.
The Descartes-Frenicle-Sorli Conjecture predicts that $k=1$.  Suppose that this Conjecture holds.
Then
$$I(q^k) = I(q) = \frac{q+1}{q} = 1+\frac{1}{q} \leq 1+\frac{1}{5}=\frac{6}{5}$$
and
$$I(n^2) = \frac{2}{I(q^k)} = \frac{2}{I(q)} \geq \frac{2}{\frac{6}{5}} = \frac{5}{3}.$$
Let
$$\frac{6}{5} = \bigg(\frac{5}{3}\bigg)^y.$$
Note that we then have that
$$I(q) \leq \frac{6}{5} = \bigg(\frac{5}{3}\bigg)^y \leq \bigg(I(n^2)\bigg)^y = \bigg(\frac{2}{I(q)}\bigg)^y$$
where
$$y = \frac{\log\bigg(\frac{6}{5}\bigg)}{\log\bigg(\frac{5}{3}\bigg)} \approx 0.356915448856724.$$
WLOG, if we assume that $k>1$ and let
$$\frac{5}{4} = \bigg(\frac{8}{5}\bigg)^z,$$
then we have that
$$I(q^k) < \frac{5}{4} = \bigg(\frac{8}{5}\bigg)^z < \bigg(I(n^2)\bigg)^z = \bigg(\frac{2}{I(q^k)}\bigg)^z$$
where
$$z = \frac{\log\bigg(\frac{5}{4}\bigg)}{\log\bigg(\frac{8}{5}\bigg)} \approx 0.474769847356948651282146696312271.$$
 A: It turns out that
$$I(q^k) < \sqrt[3]{\frac{2}{I(q^k)}}$$
implies 
$$1 + \frac{1}{q} = I(q) \leq I(q^k) < \sqrt[4]{2}$$
from which we obtain
$$q > \bigg(\sqrt[4]{2} - 1\bigg)^{-1} \approx 5.2852135$$
thereby giving
$$q \geq 13$$
since $q$ is a prime satisfying $q \equiv 1 \pmod 4$.  Thus, the implication
$$I(q^k) < \sqrt[3]{\frac{2}{I(q^k)}} \implies q \geq 13$$
holds.
If the reverse inequality
$$I(q^k) > \sqrt[3]{\frac{2}{I(q^k)}}$$
holds, then we get
$$\frac{q}{q - 1} > I(q^k) > \sqrt[4]{2}$$
which implies that
$$1 < q < \frac{\sqrt[4]{2}}{\sqrt[4]{2} - 1} \approx 6.28521$$
from which we conclude that $q = 5$.
Since $q \geq 5$ and $q$ is a prime satisfying $q \equiv 1 \pmod 4$, by the contrapositive of the last implication, we get the implication
$$q \geq 13 \implies I(q^k) < \sqrt[3]{\frac{2}{I(q^k)}}.$$
We therefore have the biconditional
$$I(q^k) < \sqrt[3]{\frac{2}{I(q^k)}} \iff q \geq 13.$$
(Note that we cannot have
$$I(q^k) = \sqrt[3]{\frac{2}{I(q^k)}}$$
as equality implies that $I(q^k) = \sqrt[4]{2}$, contradicting the fact that
$$I(q^k) = \frac{q^{k+1} - 1}{q^k (q - 1)}$$
is rational.)

Thus, to prove the inequality
  $$I(q^k) < \sqrt[3]{\frac{2}{I(q^k)}}$$
  we need to rule out $q=5$.  This is currently open, and is also unknown even if we assume the Descartes-Frenicle-Sorli Conjecture that $k=1$.

