<s>[INST] Does the inequality $I(n^2) \leq 2 - \frac{5}{3q}$ improve $I(q^k) + I(n^2) < 3$, if $q^k n^2$ is an odd perfect number with special prime $q$? Let $N = q^k n^2$ be an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
Denote the abundancy index of the positive integer $x$ by $I(x)=\sigma(x)/x$, where $\sigma(x)=\sigma_1(x)$ is the classical sum of divisors of $x$.
Here is my initial question:

Does the inequality $I(n^2) \leq 2 - \frac{5}{3q}$ lead to an improvement to the upper bound $I(q^k) + I(n^2) < 3$, if $q^k n^2$ is an odd perfect number with special prime $q$?

MOTIVATION
Here is a proof that $I(n^2) \leq 2 - \frac{5}{3q}$ holds in general.
Suppose to the contrary that $I(n^2) > 2 - \frac{5}{3q}$ is true.
Note that
$$I(n^2) = \frac{2}{I(q^k)} = \frac{2q^k (q - 1)}{q^{k+1} - 1} = 2 - 2\cdot\Bigg(\frac{q^k - 1}{q^{k+1}-1}\Bigg),$$
so that we have
$$I(n^2) > 2 - \frac{5}{3q} \iff 2 - 2\cdot\Bigg(\frac{q^k - 1}{q^{k+1}-1}\Bigg) > 2 - \frac{5}{3q} \iff \frac{5}{3q} > 2\cdot\Bigg(\frac{q^k - 1}{q^{k+1}-1}\Bigg) \iff 5q^{k+1} - 5 > 6q^{k+1} - 6q \iff 0 > q^{k+1} - 6q + 5,$$
which then implies that $k=1$.  (Otherwise, if $k > 1$, we have
$$0 > q^{k+1} - 6q + 5 \geq q^6 - 6q + 5,$$
since $k \equiv 1 \pmod 4$, contradicting $q \geq 5$.)  Now, since $k=1$, we get
$$0 > q^2 - 6q + 5 = (q - 1)(q - 5),$$
which implies that $1 < q < 5$.  This contradicts $q \geq 5$.  This concludes the proof.
Now, let $Q = 2 - \frac{5}{3q}$.
Since $I(q^k) < I(n^2)$, then we obtain
$$I(q^k) < I(n^2) \leq Q \iff (I(q^k) - Q)(I(n^2) - Q) \geq 0$$
$$\iff 2 + Q^2 = I(q^k)I(n^2) + Q^2 \geq Q(I(q^k) + I(n^2) \iff I(q^k) + I(n^2) \leq \frac{2}{Q} + Q.$$
But $\frac{2}{Q} + Q$ can be rewritten as
$$\dfrac{2}{Q} + Q = \dfrac{2}{2 - \dfrac{5}{3q}} + \Bigg(2 - \dfrac{5}{3q}\Bigg) = 3 - \Bigg(\dfrac{5}{3q} - \dfrac{5}{6q-5}\Bigg) = 3 - \dfrac{5(3q - 5)}{3q(6q - 5)} = \dfrac{54q^2 - 60q + 25}{3q(6q - 5)}.$$
Let
$$f(q) = \dfrac{54q^2 - 60q + 25}{3q(6q - 5)}.$$
Then the derivative
$$f'(q) = \dfrac{5}{3q^2} - \dfrac{30}{(6q - 5)^2}$$
is positive for $q \geq 5$.  This means that $f$ is an increasing function of $q$, which implies that
$$I(q^k) + I(n^2) \leq \dfrac{2}{Q} + Q < \lim_{q \rightarrow \infty}{f(q)} = 3.$$
FINAL QUESTION

Can we do better?  If that is not possible, can you explain why?

 [/INST] This is a partial answer.

Note that
$$I(n^2) = \dfrac{2}{I(q^k)} \leq \dfrac{2}{I(q)} = \dfrac{2q}{q+1},$$
since $k \equiv 1 \pmod 4$ implies that $k \geq 1$.
Following exactly the same steps in the original post, we obtain
$$I(q^k) < I(n^2) \leq \dfrac{2q}{q+1}.$$
After working out the details, we get
$$I(q^k) + I(n^2) \leq 3 - \Bigg(\dfrac{q - 1}{q(q + 1)}\Bigg).$$
Set $Q' = 3 - \Bigg(\dfrac{q - 1}{q(q + 1)}\Bigg)$.
Let us compute the difference $Q' - \Bigg(\dfrac{2}{Q} + Q\Bigg)$ and determine its sign for $q \geq 5$.  We obtain
$$Q' - \Bigg(\dfrac{2}{Q} + Q\Bigg) = \dfrac{5(3q - 5)}{3q(6q - 5)} - \Bigg(\dfrac{q - 1}{q(q + 1)}\Bigg) = -\dfrac{(3q - 8)(q - 5)}{3q(q + 1)(6q - 5)} \leq 0,$$
with equality occurring if and only if $q = 5$.
This shows that $Q'$ is a better upper bound for $I(q^k) + I(n^2)$ than $\dfrac{2}{Q} + Q$.

In other words, $\dfrac{2}{Q} + Q$ is a trivial upper bound for $I(q^k) + I(n^2)$.
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