(Note: This post is an offshoot of this earlier question.)
The topic of odd perfect numbers likely needs no introduction.
Denote the sum of divisors of the positive integer $x$ by $\sigma(x)$, and denote the abundancy index of $x$ by $I(x)=\sigma(x)/x$.
Euler proved that an odd perfect number $n$, if one exists, must have the form $$n = p^k m^2$$ where $p$ is the special/Euler prime satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$.
Descartes, Frenicle, and subsequently Sorli conjectured that $k=1$ always holds.
Here is my question:
Does $p \leq P$ follow from $$I(p^k)+I(m^2) \leq 3 - \bigg(\frac{p-1}{p(p+1)}\bigg),$$ if $p^k m^2$ is an odd perfect number with special prime $p$, where we set $$0 < \varepsilon = \frac{p-1}{p(p+1)}$$ and $P$ is some finite constant?
MY ATTEMPT
Notice that the inequality $$I(p^k)+I(m^2) \leq 3 - \bigg(\frac{p-1}{p(p+1)}\bigg)$$ holds in general, since $$\bigg[I(p^k) - \frac{2p}{p+1}\bigg]\bigg[I(m^2) - \frac{2p}{p+1}\bigg] \geq 0$$ follows from $$I(p^k) < I(m^2) = \frac{2}{I(p^k)} \leq \frac{2}{I(p)} = \frac{2p}{p+1}.$$
Furthermore, note that we obtain the upper bound $$\varepsilon = \frac{p-1}{p(p+1)} = \bigg(1 - \frac{1}{p}\bigg)\bigg(\frac{1}{p+1}\bigg) < \frac{1}{p + 1} \leq \frac{1}{6},$$ since $p$ is prime with $p \equiv 1 \pmod 4$ implies that $p \geq 5$.
We now compute for $p$ in terms of $\varepsilon$:
$$\varepsilon p^2 + p(\varepsilon - 1) + 1 = 0$$ $$p = \frac{(1-\varepsilon) \pm \sqrt{(1-\varepsilon)^2 - 4\varepsilon}}{2\varepsilon}$$
This gives $$p = \frac{(1-\varepsilon) \pm \sqrt{{\varepsilon}^2 - 6\varepsilon + 1}}{2\varepsilon}.$$
Quoting verbatim from [Remark 11, page 5 of The Abundancy Index of Divisors of Odd Perfect Numbers by Dris (JIS, 2012)]: Remark 11. As remarked by Joshua Zelinsky in 2005: "Any improvement on the upper bound of $3$ would have (similar) implications for all arbitrarily large primes and thus would be a very major result." (e.g. $L(p) < 2.99$ implies $p \leq 97$.) In this sense, the inequality $$\frac{57}{20} < I(p^k) + I(m^2) < 3$$ is best-possible.
I tried using WolframAlpha to get the global maxima of the two functions $$p(\varepsilon_1) = \frac{(1-\varepsilon_1) + \sqrt{{\varepsilon_1}^2 - 6{\varepsilon_1} + 1}}{2\varepsilon_1}$$ and $$p(\varepsilon_2) = \frac{(1-\varepsilon_2) - \sqrt{{\varepsilon_2}^2 - 6{\varepsilon_2} + 1}}{2\varepsilon_2},$$ but the outputs were not helpful.
Alas, this is where I get stuck.
This answer proves the following three claims :
Claim 1 : There is no $x$ such that $\dfrac{1-x - \sqrt{x^2 - 6x + 1}}{2x}\ge 5\ $ and $\ 0\lt x\le \dfrac 16$
Claim 2 : $\dfrac{1-x + \sqrt{x^2 - 6x + 1}}{2x}\ge 5\ $ and $\ 0\lt x\le \dfrac 16\iff 0\lt x\le \dfrac{2}{15}$
Claim 3 : $\displaystyle\lim_{x\to 0^+}\dfrac{1-x + \sqrt{x^2 - 6x + 1}}{2x}=+\infty$
Note that $$x^2 - 6x + 1\ge 0\quad\text{and}\quad 0\lt x\le\frac 16\iff 0\lt x\le \frac 16$$
Claim 1 : There is no $x$ such that $\dfrac{1-x - \sqrt{x^2 - 6x + 1}}{2x}\ge 5\ $ and $\ 0\lt x\le \dfrac 16$
Proof :
Suppose that there is such an $x$. Then, we have $$\begin{align}&\frac{1-x - \sqrt{x^2 - 6x + 1}}{2x}\ge 5\quad\text{and}\quad 0\lt x\le \dfrac 16 \\\\&\implies \sqrt{x^2 - 6x + 1}\le 1-11x\quad\text{and}\quad 0\lt x\le \dfrac 16 \\\\&\implies\sqrt{x^2 - 6x + 1}\le 1-11x\quad\text{and}\quad 0\lt x\le \dfrac 16\quad\text{and}\quad 0\le 1-11x \\\\&\implies x^2 - 6x + 1\le (1-11x)^2\quad\text{and}\quad 0\lt x\le \dfrac 1{11} \\\\&\implies x\ge\frac{2}{15}\quad\text{and}\quad 0\lt x\le \dfrac 1{11}\end{align}$$ which is impossible.
Claim 2 : $\dfrac{1-x + \sqrt{x^2 - 6x + 1}}{2x}\ge 5\ $ and $\ 0\lt x\le \dfrac 16\iff 0\lt x\le \dfrac{2}{15}$
Proof :
$$\begin{align}&\dfrac{1-x + \sqrt{x^2 - 6x + 1}}{2x}\ge 5\quad\text{and}\quad 0\lt x\le \dfrac 16 \\\\&\iff \sqrt{x^2 - 6x + 1}\ge 11x-1\quad\text{and}\quad 0\lt x\le \dfrac 16 \\\\&\iff 0\lt x\le\frac{1}{11}\quad\text{or}\quad\bigg(x^2-6x+1\ge (11x-1)^2\quad\text{and}\quad \frac{1}{11}\lt x\le\frac 16\bigg) \\\\&\iff 0\lt x\le\frac{1}{11}\quad\text{or}\quad\bigg(0\le x\le\frac{2}{15}\quad\text{and}\quad \frac{1}{11}\lt x\le\frac 16\bigg) \\\\&\iff 0\lt x\le\frac{1}{11}\quad\text{or}\quad\frac{1}{11}\lt x\le\frac{2}{15} \\\\&\iff 0\lt x\le\frac{2}{15}\end{align}$$
Claim 3 : $\displaystyle\lim_{x\to 0^+}\dfrac{1-x + \sqrt{x^2 - 6x + 1}}{2x}=+\infty$
Proof : Since $\displaystyle\lim_{x\to 0^+}(1-x + \sqrt{x^2 - 6x + 1})=2$, we get $$\displaystyle\lim_{x\to 0^+}\dfrac{1-x + \sqrt{x^2 - 6x + 1}}{2x}=+\infty$$