(Preamble: This question is an offshoot of this earlier post.)
Denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$, and the abundancy index of $x$ by $I(x)=\sigma(x)/x$.
Suppose that there exists a (hypothetical) odd perfect number $N = p^k m^2$ with special prime $p$ satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$.
Since $p$ is prime, we have the formula and corresponding (sharp?) upper bound $$I(p^k)=\dfrac{\sigma(p^k)}{p^k}=\dfrac{p^{k+1}-1}{p^k (p-1)}<\dfrac{p^{k+1}}{p^k (p - 1)}=\dfrac{p}{p-1}.$$ Moreover, since $p \equiv 1 \pmod 4$ holds, then $p \geq 5$. This implies that $$I(p^k)<\dfrac{p}{p-1} \leq \dfrac{5}{4}.$$ We therefore have $$I(m^2)=\dfrac{2}{I(q^k)} > \dfrac{8}{5},$$ since the abundancy index function $I$ is multiplicative.
Here is my question:
Can I then say that $$I(m^2) > \dfrac{8}{5} \geq I(m),$$ since $I(m^2) > I(m)$ holds for all positive integers $m$?
MY ATTEMPT
I was thinking that I could use the fact that $$\inf{I(m^2)} = \dfrac{8}{5}$$ and then "deduce" as follows: $$I(m) < I(m^2)$$ This implies (?) that $$I(m) \leq \sup{I(m)} \leq \inf{I(m^2)} = \dfrac{8}{5}.$$
MY SECOND ATTEMPT
We have $$\bigg(\dfrac{2(p-1)}{p}\bigg)^{\dfrac{\ln(4/3)}{\ln(13/9)}} < \bigg(I(m^2)\bigg)^{\dfrac{\ln(4/3)}{\ln(13/9)}} < I(m),$$ as communicated by Pascal Ochem in an e-mail dated April 17, 2013.
So if $I(m) \leq 8/5$, then we have the inequality $$\bigg(\dfrac{2(p-1)}{p}\bigg)^{\dfrac{\ln(4/3)}{\ln(13/9)}} < \dfrac{8}{5},$$ which results in the real solutions $$p = 1$$ or $$1 < p < \dfrac{2\exp\Bigg(\frac{\ln(8/5)\ln(9/13)}{\ln(4/3)}\Bigg)}{2\exp\Bigg(\frac{\ln(8/5)\ln(9/13)}{\ln(4/3)}\Bigg) - 1} \approx 11.3334$$
Since $p$ is the special prime (satisfying $p \equiv 1 \pmod 4$), then $p = 5$.
I therefore predict that:
CONJECTURE: If $p^k m^2$ is an odd perfect number with special prime $p$, then $p = 5$.
Alas, this is where I get stuck, as I have no proof of this conjecture.