On improving the upper bound $I(m^2) \leq \frac{2p}{p+1}$, if $p^k m^2$ is an odd perfect number with special prime $p$ (Preamble: This question is an offshoot of this earlier MSE post.)
Let $p^k m^2$ be an odd perfect number (OPN) with special prime $p$ satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$.
Denote the classical 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 abundancy index of $x$ by $I(x)=\sigma(x)/x$.
Note that it is trivial to prove that
$$\frac{p+1}{p} \leq I(p^k) < \frac{p}{p-1}$$
from which we obtain
$$\frac{2(p-1)}{p} < I(m^2) = \frac{2}{I(p^k)} \leq \frac{2p}{p+1}.$$
This implies that
$$\frac{2}{p+1} \leq \frac{D(m^2)}{m^2} < \frac{2}{p}.$$
Taking reciprocals, multiplying by $2$, and subtracting $1$, we get
$$p-1 < \frac{\sigma(m^2)}{D(m^2)} \leq p,$$
where we note that $(2/p) < p-1$.
Now consider the quantity
$$\bigg(\dfrac{D(m^2)}{m^2} - \dfrac{2}{p}\bigg)\bigg(\dfrac{\sigma(m^2)}{D(m^2)} - \dfrac{2}{p}\bigg).$$
This quantity is negative.  Thus, we obtain
$$I(m^2) + \bigg(\dfrac{2}{p}\bigg)^2 < \dfrac{2}{p}\bigg(\dfrac{D(m^2)}{m^2} + \dfrac{\sigma(m^2)}{D(m^2)}\bigg) = \dfrac{2}{p}\Bigg(\bigg(2-I(m^2)\bigg) + \bigg(\dfrac{I(m^2)}{2-I(m^2)}\bigg)\Bigg).$$
Now, let $z_1=I(m^2)$.  Then we have the inequality
$$z_1 + \bigg(\dfrac{2}{p}\bigg)^2 < \dfrac{2}{p}\Bigg(\bigg(2-z_1\bigg) + \bigg(\dfrac{z_1}{2-z_1}\bigg)\Bigg),$$
from which we obtain
$$\dfrac{2(p-1)}{p} < z_1=I(m^2) < 2$$
using WolframAlpha.
Here is my inquiry:

QUESTION: In this closely related MSE question, we were able to derive the improved lower bound
$$\frac{2(p-1)}{p}+\frac{1}{pm^2}<I(m^2).$$
Can we similarly derive an improved upper bound for $I(m^2)$, that is hopefully better than
$$I(m^2) \leq \frac{2p}{p+1}?$$
If we cannot, then can you explain why?


MY ATTEMPT
Consider the quantity
$$\bigg(\dfrac{D(m^2)}{m^2} - \dfrac{2}{p+1}\bigg)\bigg(\dfrac{\sigma(m^2)}{D(m^2)} - \dfrac{2}{p+1}\bigg).$$
This quantity is nonnegative.  Thus, we obtain
$$I(m^2) + \bigg(\dfrac{2}{p+1}\bigg)^2 \geq \dfrac{2}{p+1}\bigg(\dfrac{D(m^2)}{m^2} + \dfrac{\sigma(m^2)}{D(m^2)}\bigg) = \dfrac{2}{p+1}\Bigg(\bigg(2-I(m^2)\bigg) + \bigg(\dfrac{I(m^2)}{2-I(m^2)}\bigg)\Bigg).$$
Now, let $z_2 = I(m^2)$.  Then we have the inequality
$$z_2 + \bigg(\dfrac{2}{p+1}\bigg)^2 \geq \dfrac{2}{p+1}\Bigg(\bigg(2-z_2\bigg) + \bigg(\dfrac{z_2}{2-z_2}\bigg)\Bigg),$$
from which we obtain
$$\dfrac{4}{p+3} \leq z_2=I(m^2) \leq \dfrac{2p}{p+1},$$
using WolframAlpha, which does not improve on the previous known bounds for $I(m^2)$.

This is an experimental attempt - my apologies for any silly mistakes.

Since we have
$$I(m^2) \leq \dfrac{2p}{p+1}$$
and because $p^k m^2$ is perfect, then we have
$$I(p^k)I(m^2) = I(p^k m^2) = 2$$
where we have used the fact that the abundancy index function is multiplicative.
Hence, we obtain (via iteration)
$$\text{First iteration: } I(m^2) \leq I(p^k)I(m^2)\cdot\dfrac{p}{p+1} \leq 2I(p^k)\cdot\dfrac{p^2}{(p+1)^2} = \dfrac{2p^2\Bigg(p^{k+1} - 1\Bigg)}{p^k (p-1)(p+1)^2}$$
$$\text{Second iteration: } I(m^2) \leq 2I(p^k)\cdot\dfrac{p^2}{(p+1)^2}$$ $$= \bigg(I(p^k)\bigg)^2 {I(m^2)} \cdot\dfrac{p^2}{(p+1)^2} \leq 2\bigg(I(p^k)\bigg)^2 \cdot\dfrac{p^3}{(p+1)^3}$$
$$\ldots$$
$$\ldots$$
$$\ldots$$
by recursively replacing $2$ with $I(p^k)I(m^2)$ and then bounding $I(m^2)$ from above by $2p/(p+1)$.
Note that, at the $n^{\text{th}}$ iteration, we have the inequality
$$I(m^2) \leq 2\bigg(I(p^k)\bigg)^n \cdot \dfrac{p^{n+1}}{(p+1)^{n+1}}.$$
Repeating the process ad infinitum, we get
$$I(m^2) \leq \lim_{n \rightarrow \infty}{\Bigg(2\bigg(I(p^k)\bigg)^n \cdot \dfrac{p^{n+1}}{(p+1)^{n+1}}\Bigg)} = \lim_{n \rightarrow \infty}{\Bigg(\dfrac{2\bigg(I(p^k)\bigg)^n}{\dfrac{(p+1)^{n+1}}{p^{n+1}}}\Bigg)},$$
a limit which is of the indeterminate form
$$\dfrac{+\infty}{+\infty}.$$
Hence, we may apply L'Hôpital's rule, by separately considering
$$h(p) = \Bigg(\dfrac{2\bigg(I(p^k)\bigg)^n}{\dfrac{(p+1)^{n+1}}{p^{n+1}}}\Bigg)$$
as a function of $p$, and
$$j(k) = \Bigg(\dfrac{2\bigg(I(p^k)\bigg)^n}{\dfrac{(p+1)^{n+1}}{p^{n+1}}}\Bigg)$$
as a function of $k$.
(Also, note that there is "currently no L'Hôpital's rule for multiple variable limits".  The closest thing that I could find on arXiv is this preprint by Gary R. Lawlor.)
I will stop here for the time being.
Update: (October 13, 2021 - 11:37 AM) - Manila time Since the denominator of $j(k)$ is strictly a function of $p$ and $n$, then there is no need to consider $j(k)$.
Working on $h(p)$ now, will post an update in a bit.
Update: (October 13, 2021 - 12:27 PM) - Manila time Cross-posted this answer as a question to MO, since the computations involved are somewhat tedious and complicated.
