The topic of odd perfect numbers likely needs no introduction.
Let $\sigma(x)=\sigma_1(x)$ denote the classical sum of divisors of the positive integer $x$, and let $I(x)=\sigma(x)/x$ denote the abundancy index of $x$. 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 $\gcd(p,m)=1$.
In this paper (page $112$), it is stated that:
An improvement to the currently known upper bound of $I(m) < 2$ will be considered a major breakthrough. In the sequel (?), a viable approach towards improving the inequality $I(m) < 2$ will be presented, which may necessitate the use of ideas from the paper by Ward.
I revisited the argument(s) in the sequel, and there was no mention of a specific connection to Ward's ideas.
So here I go now: Suppose $10$ divides $m^2$. (This is actually a false assumption, but oh well.)
Then $$\frac{9}{5} = I(10) \leq I(m^2).$$
Now, here is where it becomes very interesting. Take the contrapositive of the last statement.
Suppose that $I(m^2) < 9/5$. Then $10$ does not divide $m^2$.
Note that the conclusion "$10$ does not divide $m^2$" of the contrapositive is true (since $m$ is odd).
Notice that, if $I(m^2) < 9/5$ were true, then we would have $$\dfrac{2(p-1)}{p} < I(m^2) < \dfrac{9}{5} \iff 10(p - 1) < 9p \iff p < 10 \implies p = 5$$ and $$I(m) < I(m^2) < \dfrac{9}{5}.$$
Here is my:
INITIAL QUESTION: Since it is known that $I(m^2) > 8/5$, will it be possible to prove that the upper bound $I(m^2) < 9/5$ holds?
MY ATTEMPT
Suppose that $I(m^2) > 9/5$. But we have $$\dfrac{2p}{p+1} \geq I(m^2) > \dfrac{9}{5},$$ which implies that $$10p > 9p + 9$$ $$p > 9$$ $$p \geq 13 \implies I(p^k) < \dfrac{p}{p - 1} \leq \dfrac{13}{12} \implies I(m^2) = \dfrac{2}{I(p^k)} > \dfrac{24}{13} > \dfrac{9}{5}.$$
Hence, we have the biconditional $$I(m^2) > \dfrac{9}{5} \iff p \geq 13,$$ so that, in order to prove $I(m^2) < 9/5$, we need to rule out $p \geq 13$.
FINAL QUESTION: Has anyone actually conjectured in the literature that $p = 5$? (Pace Nielsen says in this MO answer that, even if $p = 5$ and $k = 1$, there are no known effective bounds on $N$.)