If $p^k m^2$ is an odd perfect number, then is there a constant $D$ such that $\frac{\sigma(m^2)}{p^k} > \frac{m^2 - p^k}{D}$? (Note:  This question is an offshoot of this closely related one.)
Denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$.
The topic of odd perfect numbers likely needs no introduction.
The initial question is as is in the title:

If $p^k m^2$ is an odd perfect number with special prime $p$, then is there a constant $D$ such that $$\frac{\sigma(m^2)}{p^k} > \frac{m^2 - p^k}{D}?$$

(Note that the special prime $p$ satisfies $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$.)
MY ATTEMPT
Since $\gcd(p^k,\sigma(p^k))=1$, we know that
$$\frac{\sigma(m^2)}{p^k}=\frac{2m^2}{\sigma(p^k)}=\frac{D(m^2)}{s(p^k)},$$
where $D(x)=2x-\sigma(x)$ is the deficiency of $x$, and $s(x)=\sigma(x)-x$ is the aliquot sum of $x$.
Now, let $D > 0$ be a constant such that
$$\frac{\sigma(m^2)}{p^k} > \frac{m^2 - p^k}{D}.$$
This is equivalent to
$$D > \frac{p^k(m^2 - p^k)}{\sigma(m^2)} = \frac{p^k \sigma(p^k) (m^2 - p^k)}{\sigma(p^k)\sigma(m^2)} = \frac{p^k \sigma(p^k) (m^2 - p^k)}{2p^k m^2}$$
$$= \frac{\sigma(p^k)}{2} - \frac{p^k \sigma(p^k)}{2 m^2} = \frac{\sigma(p^k)}{2} - \frac{p^k \cdot p^k}{\sigma(m^2)}.$$
Using mathlove's answer to a closely related question, we have the lower bound
$$\frac{\sigma(m^2)}{p^k} \geq 3^3 \times 5^3 = 3375.$$
Hence, we have
$$D > \frac{\sigma(p^k)}{2} - \frac{p^k \cdot p^k}{\sigma(m^2)} \geq \frac{\sigma(p^k)}{2} - \frac{p^k}{3375} = \frac{p^{k+1} - 1}{2(p - 1)} - \frac{p^k}{3375} = \frac{3373p^{k+1} + 2p^k - 3375}{6750(p - 1)}.$$
Let
$$f(k) = \frac{3373p^{k+1} + 2p^k - 3375}{6750(p - 1)}.$$
Then the first derivative
$$\frac{\partial f}{\partial k} = \frac{(3373p + 2)p^k \log(p)}{6750(p - 1)}$$
is positive for $p \geq 5$.  This implies that $f$ is an increasing function of $k$.
Therefore, since $k \geq 1$, we obtain
$$D > \frac{\sigma(p^k)}{2} - \frac{p^k \cdot p^k}{\sigma(m^2)} \geq f(k) \geq f(1) = \frac{3373p}{6750} + \frac{1}{2}.$$
But we know that $p$ is at least $5$, since $p$ is the special prime satisfying $p \equiv 1 \pmod 4$.  Therefore, we have
$$D > \frac{\sigma(p^k)}{2} - \frac{p^k \cdot p^k}{\sigma(m^2)} \geq f(k) \geq f(1) = \frac{3373p}{6750} + \frac{1}{2} \geq \frac{2024}{675} = 2.9985\overline{185}.$$
Here are my final questions:

Does this mean that we can take $D=3$ in the inequality
$$\frac{\sigma(m^2)}{p^k} > \frac{m^2 - p^k}{D}?$$
If $D=3$ does not work, then what value of $D$ works?


This is a partial answer.
If $D=3$, then $p=5$ and $k=1$ both hold.
Suppose that $D=3$.
$$\frac{\sigma(m^2)}{p^k} > \frac{m^2 - p^k}{3}$$
$$2 > I(m^2)=\frac{\sigma(m^2)}{m^2} > \frac{p^k \bigg(1 - \dfrac{p^k}{m^2}\bigg)}{3}$$
$$\implies 6 > p^k \bigg(1 - \frac{p^k}{m^2}\bigg) \geq p^k \bigg(1 - \frac{2}{3375}\bigg) = \frac{3373p^k}{3375}$$
since
$$\frac{\sigma(m^2)}{p^k} \geq 3375 \implies \frac{p^k}{m^2} \leq \frac{2}{3375}.$$
But the inequality
$$\frac{3373p^k}{3375} < 6$$
only holds when $p = 5$ and $k = 1$.   QED
Hence, we may not take $D=3$, as it will not make the inequality hold in general.
