(Note: This is a continuation of this earlier question.)
Let $N = q^k n^2$ be an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$. (In particular, we know that $q \geq 5$.)
Denote the sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$, the deficiency of $x$ by $D(x)=2x-\sigma(x)$, and the sum of aliquot/proper divisors of $x$ by $s(x)=\sigma(x)-x$. Finally, denote the abundancy index of $x$ by $I(x)=\sigma(x)/x$.
In this accepted answer to a closely related question, it was proved that $$\frac{\sigma(n^2)}{q^k} \leq \frac{n^2 - q^k}{\frac{188}{63}}.$$
Since we also have, in general, the following equation and inequalities $$\frac{\sigma(n^2)}{q^k} = \frac{D(n^2)}{s(q^k)} \leq \frac{\sigma(n^2)}{q} \leq D(n^2),$$ here is my:
QUESTION: If $N=q^k n^2$ is an odd perfect number with special prime $q$, is the following inequality true in general? $$D(n^2) \leq \frac{n^2 - q^k}{\frac{188}{63}}$$
MY ATTEMPT
The inequality under consideration in the question is equivalent to $$376n^2 - 188\sigma(n^2) \leq 63n^2 - 63q^k.$$ Suppose to the contrary that it is false. This implies that $$I(n^2) < \frac{313}{188} + \frac{63q^k}{188n^2}.$$ But we have the estimate $$\frac{q^k}{n^2} < \frac{2}{315}$$ by Broughan, Delbourgo, and Zhou (2013), which implies that $$\frac{2(q-1)}{q} < I(n^2) < \frac{313}{188} + \frac{63q^k}{188n^2} < \frac{313}{188} + \frac{63}{188}\cdot\frac{2}{315} = \frac{1567}{940},$$ from which it follows that $$q < \frac{1880}{313} \approx 6.00639.$$ Since $q$ is a prime which satisfies $q \equiv 1 \pmod 4$, then $q=5$. Alas, this is where I get stuck.
