Preamble: This question is an offshoot of this earlier post. (This inquiry has likewise been cross-posted to MO last June $10, 2022$.)
Let $N = q^k n^2$ be an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
Define the abundancy index $$I(x)=\frac{\sigma(x)}{x}$$ where $\sigma(x)$ is the classical sum of divisors of $x$.
Since $q$ is prime, we have the bounds $$\frac{q+1}{q} \leq I(q^k) < \frac{q}{q-1},$$ which implies, since $N$ is perfect, that $$\frac{2(q-1)}{q} < I(n^2) = \frac{2}{I(q^k)} \leq \frac{2q}{q+1}.$$
By considering the negative product $$\bigg(I(q^k) - \frac{2(q-1)}{q}\bigg)\bigg(I(n^2) - \frac{2(q-1)}{q}\bigg) < 0,$$ since we obviously have $$\frac{q}{q-1} < \frac{2(q-1)}{q},$$ then after some routine algebraic manipulations, we arrive at the lower bound $$I(q^k) + I(n^2) > 3 - \frac{q-2}{q(q-1)} = \frac{3q^2 - 4q + 2}{q(q - 1)}.$$
Note that, since $q$ is a prime satisfying $q \equiv 1 \pmod 4$, then $q \geq 5$, so that $$f(q) = \frac{3q^2 - 4q + 2}{q(q - 1)}$$ is increasing, since $$f'(q) = \frac{q(q - 4) + 2}{\bigg(q(q - 1)\bigg)^2} > 0.$$ This means that $$I(q^k) + I(n^2) > f(q) \geq f(5) = \frac{57}{20}.$$
By this answer to a closely related MO question, we have the improved bounds: $$I(q^k) < \frac{q}{q - 1} < \frac{2(q - 1)}{q} + \frac{1}{qn^2} < I(n^2).$$
Using my method, this implies that the product $$(I(q^k) - y)(I(n^2) - y) < 0$$ is negative, where $$y = \dfrac{2(q-1)}{q} + \dfrac{1}{qn^2}.$$ After some algebraic simplifications, we get $$I(q^k) + I(n^2) > \dfrac{3q^2 - 4q + 2}{q(q - 1)} - \dfrac{q}{(q - 1)(2qn^2 - 2n^2 + 1)} + \dfrac{1}{qn^2}.$$ But $$- \dfrac{q}{(q - 1)(2qn^2 - 2n^2 + 1)} + \dfrac{1}{qn^2} = \dfrac{qn^2 (q - 4) + q + 2n^2 - 1}{qn^2 (q - 1)(2n^2 (q - 1) + 1)} > 0,$$ since $q \geq 5$.
So the inequality $$I(q^k) + I(n^2) > \dfrac{3q^2 - 4q + 2}{q(q - 1)} - \dfrac{q}{(q - 1)(2qn^2 - 2n^2 + 1)} + \dfrac{1}{qn^2} = g_1(q,n)$$ is unconditionally true, which would mean that the new lower bound $g_1(q,n)$ for $I(q^k) + I(n^2)$ improves on the old $f(q)$. Note that we were able to prove this analytically.
Now consider the expression $$g_1(q,n) - \frac{57}{20} = \Bigg(\dfrac{3q^2 - 4q + 2}{q(q - 1)} - \frac{57}{20}\Bigg) - \dfrac{q}{(q - 1)(2qn^2 - 2n^2 + 1)} + \dfrac{1}{qn^2} = a_1 + a_2,$$ where $$a_1 = \dfrac{3q^2 - 4q + 2}{q(q - 1)} - \frac{57}{20} \geq 0$$ (by our previous considerations), and $$a_2 = - \dfrac{q}{(q - 1)(2qn^2 - 2n^2 + 1)} + \dfrac{1}{qn^2} = \dfrac{qn^2 (q - 4) + q + 2n^2 - 1}{qn^2 (q - 1)(2n^2 (q - 1) + 1)} > 0.$$
We therefore conclude that $$I(q^k) + I(n^2) > g_1(q,n) > \frac{57}{20},$$ so that we do appear to have gotten an improved lower bound for $I(q^k) + I(n^2)$, better than $57/20$.
By Lemma II.2, page 1 of this conference paper, "any further improvement to the lower bound of $57/20$ for $I(q^k) + I(n^2)$ would be equivalent to showing that there are no odd perfect numbers of the form $5n^2$, which would be a very major result".
Here are my:
QUESTIONS
(1) Is this proof correct?
(2) If the answer to Question (1) is NO, can the proof be mended so as to produce a logically valid argument?
