If $p^k m^2$ is an odd perfect number with special prime $p$, then which is larger: $D(p^k)$ or $D(m^2)$? Denote the classical sum of divisors of the positive integer $x$ to be $\sigma(x)=\sigma_1(x)$.  Denote the abundancy index of $x$ by $I(x)=\sigma(x)/x$.  Finally, denote the deficiency of $x$ by $D(x)=2x-\sigma(x)$.
The topic of odd perfect numbers likely needs no introduction.
Let $p^k m^2$ be an odd perfect number with special prime $p$, satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$.
Here is my question:

If $p^k m^2$ is an odd perfect number with special prime $p$, then which is larger: $D(p^k)$ or $D(m^2)$?

MY ATTEMPT
Since $p$ is prime, we have
$$1<I(p^k)=\frac{\sigma(p^k)}{p^k}=\frac{p^{k+1}-1}{p^k (p - 1)}<\frac{p^{k+1}}{p^k (p - 1)}=\frac{p}{p-1}.$$
Now, because $p$ is a prime satisfying $p \equiv 1 \pmod 4$, we have the lower bound $p \geq 5$, whereupon we obtain the upper bound
$$I(p^k)<\frac{p}{p-1} \leq \frac{5}{4}.$$
Note that, since $p^k m^2$ is perfect and $\gcd(p,m)=1$, then we have
$$2=I(p^k m^2)=I(p^k)I(m^2) \iff I(m^2) = \frac{2}{I(p^k)}.$$
This implies that we have the lower bound
$$I(m^2) > \frac{2}{(5/4)} = \frac{8}{5},$$
from which we finally get
$$1 < I(p^k) < \frac{5}{4} < \frac{8}{5} < I(m^2) < 2,$$
since $m^2$ is a proper factor of the perfect number $p^k m^2$, and is therefore deficient.
In particular, we have
$$0 < 2 - I(m^2) < 2 - I(p^k) < 1$$
$$0 < \frac{D(m^2)}{m^2} < \frac{D(p^k)}{p^k} < 1$$
$$0 < D(m^2) < \frac{m^2}{p^k}\cdot{D(p^k)} < m^2.$$
Alas, this is where I get stuck.

Too long to comment :
One can prove that
$$D(m^2)\lt D(p^k)\iff m^2\lt \frac{(p^{k+1}-1)(p^{k+1}-2p^k+1)}{2(p-1)(p^k-1)}$$
Proof :
We have
$$\begin{align}&D(p^k)-D(m^2)
\\\\&=2p^k-\sigma(p^k)-2m^2+\sigma(m^2)
\\\\&=2p^k-\sigma(p^k)-2m^2+\frac{2p^km^2}{\sigma(p^k)}
\\\\&=2p^k-\frac{p^{k+1}-1}{p-1}-2m^2+\frac{2p^km^2(p-1)}{p^{k+1}-1}
\\\\&=\frac{(p-1)(p^{k+1}-1)(2p^k-2m^2)-(p^{k+1}-1)^2+2p^km^2(p-1)^2}{(p-1)(p^{k+1}-1)}
\\\\&=\frac{ p^{2 k + 2}- 2 p^{2 k + 1}  + 2 p^k - 1-2(p^{k + 1} - p^k  - p +1)m^2}{(p-1)(p^{k+1}-1)}
\\\\&=\frac{(p^{k+1}-1)(p^{k+1}-2p^k+1)-2(p-1)(p^k-1)m^2}{(p-1)(p^{k+1}-1)}\end{align}$$
from which we get
$$\begin{align}D(m^2)\lt D(p^k)&\iff \frac{(p^{k+1}-1)(p^{k+1}-2p^k+1)-2(p-1)(p^k-1)m^2}{(p-1)(p^{k+1}-1)}\gt 0
\\\\&\iff (p^{k+1}-1)(p^{k+1}-2p^k+1)\gt 2(p-1)(p^k-1)m^2
\\\\&\iff m^2\lt \frac{(p^{k+1}-1)(p^{k+1}-2p^k+1)}{2(p-1)(p^k-1)}\end{align}$$
