Let $\sigma(N)$ denote the sum of the divisors of the natural number $N$. If $\sigma(N)=2N$ and $N$ is odd, then $N$ is called an odd perfect number. Denote the abundancy index $I$ of $N$ as $I(N)=\sigma(N)/N$.
Euler proved that every odd perfect number $N$ has to have the form $N=q^k n^2$ where $q$ is prime with $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
In this paper, Dris proved that we have the bounds $$L(q) = \frac{3q^2 - 4q + 2}{q(q-1)} = 3 - \frac{q-2}{q(q-1)} < I_{+} \leq 3 - \frac{q-1}{q(q+1)} = \frac{3q^2 + 2q + 1}{q(q+1)} = U(q)$$ for the sum of the abundancy indices $I_{+}=I(q^k)+I(n^2)$. In particular, if $I_{+} \leq 2.99$, then we will have $L(q) < 2.99$, so that (using WolframAlpha) we get $$q < \frac{101 + \sqrt{9401}}{2} \approx 98.9794,$$ (since $q$ satisfies $q \geq 5$) from which we obtain $q \leq 97$ (since $q$ is prime and $q \equiv 1 \pmod 4$).
Here is the problem:
PROBLEM STATEMENT
Does the converse hold? That is, if $N=q^k n^2$ is an odd perfect number, does $q \leq 97$ imply that $I(q^k) + I(n^2) \leq 2.99$?
MY ATTEMPT
Since $N=q^k n^2$ is perfect and $\gcd(q,n)=1$, then $I(q^k)I(n^2)=2$, from which we have $$I(n^2) = \frac{2}{I(q^k)} \leq \frac{2}{I(q)} = \frac{2q}{q+1} = \frac{2}{1 + \frac{1}{q}}.$$
By assumption, $q \leq 97$, so that we get $$\frac{1}{q} \geq \frac{1}{97} \implies 1 + \frac{1}{q} \geq \frac{98}{97} \implies \frac{2}{1 + \frac{1}{q}} \leq \frac{97}{49} \approx 1.97959\ldots$$
However, this does not match up with the known upper bound $$I(q^k) < \frac{5}{4} = 1.25.$$
ANOTHER APPROACH
Assume that $q \leq 97$, and suppose to the contrary that $$I(q^k) + I(n^2) > 2.99.$$
This means that $U(q) > 2.99$, so that (by using WolframAlpha again) we obtain $$q > \frac{99 + \sqrt{9401}}{2} \approx 97.9794,$$ contradicting $q \leq 97$.
Here is my question:
Is this proof in the ANOTHER APPROACH section correct?
(This is for the sake of posting an answer to an unanswered question.)
Yes, the proof in the ANOTHER APPROACH section is correct.
In fact, the biconditional so presented here can be generalized as detailed in this MSE post.