<s>[INST] Does $k=1$ follow from $I(5^k)+I(m^2) \leq \frac{43}{15}$, if $p^k m^2$ is an odd perfect number with special prime $p=5$? 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 initial question:

Does $k=1$ follow from
$$I(5^k)+I(m^2) \leq \frac{43}{15},$$
if $I(x)$ is the abundancy index of $x$ and $p^k m^2$ is an odd perfect number with special prime $p=5$?

MY ATTEMPT
Since $n = p^k m^2$ is perfect and $I$ is multiplicative, then we have
$$I(m^2) = \frac{2}{I(p^k)}.$$
But $p \mid p^k$.  In particular, $I(p) \leq I(p^k)$.  This implies that we have the upper bound
$$I(m^2) = \frac{2}{I(p^k)} \leq \frac{2}{I(p)} = \frac{2p}{p+1}.$$
In particular, if $p=5$, then we obtain
$$I(m^2) \leq \frac{5}{3}.$$
Hereinafter, we shall assume that $p=5$.
Now, consider the product
$$\bigg(I(p^k) - \frac{5}{3}\bigg)\bigg(I(m^2) - \frac{5}{3}\bigg).$$
It can be proven that $I(p^k) < I(m^2)$.  (For a proof, see [Dris (2012)].)  Thus, this product is nonnegative (since $I(m^2) \leq \frac{5}{3}$), whereupon we get
$$\bigg(I(p^k) - \frac{5}{3}\bigg)\bigg(I(m^2) - \frac{5}{3}\bigg) \geq 0 \implies I(p^k)I(m^2) + \bigg(\frac{5}{3}\bigg)^2 \geq \frac{5}{3}\cdot\bigg(I(p^k) + I(m^2)\bigg)$$
$$\implies 2 + \bigg(\frac{5}{3}\bigg)^2 \geq \frac{5}{3}\cdot\bigg(I(p^k) + I(m^2)\bigg) \implies I(p^k) + I(m^2) \leq \frac{6}{5} + \frac{5}{3} = \frac{43}{15}.$$
Now, we compute an exact expression for
$$I(p^k) + I(m^2) = I(p^k) + \frac{2}{I(p^k)}$$
when $p=5$.  We obtain
$$I(5^k) + \frac{2}{I(5^k)} = \frac{(5^{k+1} - 1)^2 + {32}\cdot{5^{2k}}}{5^k \cdot {4} (5^{k+1} - 1)} = \frac{57 \cdot {5^{2k}} - 2 \cdot {5^{k+1}} + 1}{4 \cdot {5^{2k+1}} - 4 \cdot {5^k}}.$$
From the same paper cited above, we have the lower bound
$$\frac{57}{20} < I(p^k) + I(m^2)$$
so that collectively we have
$$\frac{57}{20} < I(5^k) + \frac{2}{I(5^k)} = \frac{57 \cdot {5^{2k}} - 2 \cdot {5^{k+1}} + 1}{4 \cdot {5^{2k+1}} - 4 \cdot {5^k}} \leq \frac{43}{15},$$
from which we obtain the (trivial) lower bound
$$k \geq 1,$$
per this WolframAlpha computation.
Here is my follow-up question:

Why does the condition
$$I(5^k) + \frac{2}{I(5^k)} \in \bigg(\frac{57}{20},\frac{43}{15}\bigg]$$
not result in a nontrivial bound for $k$?

 [/INST] On OP's request, I am converting my comment into an answer.
I'm not sure if I understand your follow-up question well, but we have
$$\begin{align}\frac{\mathrm d}{\mathrm dk}\bigg(I(5^k)+\frac{2}{I(5^k)}\bigg)&=-\frac{ 7\cdot 5^{2 k} + 2\cdot 5^{k + 1} - 1}{4\cdot 5^k (5^{k + 1} - 1)^2}\log 5\lt 0
\\\\\lim_{k\to\infty}\bigg(I(5^k)+\frac{2}{I(5^k)}\bigg)&=\lim_{k\to\infty} \frac{57  -\frac{ 2}{5^{k-1}} + \frac{1}{5^{2k}}}{20 - \frac{4}{5^k}}=\frac{57}{20}
\\\\I(5)+\frac{2}{I(5)}&=\frac{57 \cdot {5^{2}} - 2 \cdot {5^{2}} + 1}{4 \cdot {5^{3}} - 4 \cdot {5}}=\frac{43}{15}\end{align}$$
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