Denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$, and denote the abundancy index of $x$ by $I(x)=\sigma(x)/x$.
Let $N = q^k n^2$ be an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
It is known that $$\dfrac{57}{20} < I(q^k) + I(n^2) < 3.$$
Note that $$I(q^k) + I(n^2) = I(q^k) + \dfrac{2}{I(q^k)}$$ and $$I(q^k) + I(n^2) = I(n^2) + \dfrac{2}{I(n^2)},$$ so that we have $$\dfrac{57}{20} < I(n^2) + \dfrac{2}{I(n^2)} < 3.$$
Here is my:
QUESTION: Does this inequality imply that $n^2$ is bounded?
MY ATTEMPT
I have $$n^2 < q^k n^2 = N < 2^{4^{\omega(N)}},$$ where $\omega(N)$ is the number of distinct prime factors of $N$. (Note that the upper bound for $N$ is due to Nielsen.)
Hence, $n^2$ is bounded if $\omega(N)$ is bounded.
Alas, this is where I get stuck.