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 $x$ as $I(x)=\sigma(x)/x$ where $\sigma(x)=\sigma_1(x)$ is the classical sum of divisors of $x$. (For example, when $M$ is perfect, we have $\sigma(M)=2M$ and $I(M)=\sigma(M)/M=2$. The smallest example of an even perfect number is $6$. It is currently unknown if there are any odd perfect numbers.)
Fix a positive integer $a$.
We have $$k \leq a \iff I(q^k) \leq I(q^a) \iff \frac{2}{I(q^a)} \leq \frac{2}{I(q^k)}=I(n^2)$$ $$\iff I(q^k) \leq I(q^a) < \frac{2}{I(q^a)} \leq I(n^2),$$ since $$I(q^a) = \frac{q^{a+1} - 1}{q^a (q - 1)} < \frac{q}{q-1} \leq \frac{5}{4} < \sqrt{2}.$$ Therefore, we get $$k \leq a \iff \Bigg[I(q^k) - I(q^a)\Bigg]\Bigg[I(n^2) - I(q^a)\Bigg] \leq 0$$ $$\iff 2 + (I(q^a))^2 \leq I(q^a)\Bigg[I(q^k) + I(n^2)\Bigg].$$ We thus obtain the following proposition:
PROPOSITION If $N=q^k n^2$ is an odd perfect number with special prime $q$, then $k \leq a$ holds if and only if $$I(q^a) + \frac{2}{I(q^a)} \leq I(q^k) + I(n^2).$$
ALTERNATIVE PROOF OF THE PROPOSITION Note that the same proposition can be proved by using the fact that $$f(k) = I(q^k) + I(n^2) = I(q^k) + \frac{2}{I(q^k)}$$ is a decreasing function of $k$. (See mathlove's answer to a closely related question.)
But $$I(q^a) + \frac{2}{I(q^a)} = \frac{q^{a+1} - 1}{q^a (q - 1)} + \frac{2q^a(q - 1)}{q^{a+1} - 1} = \frac{3q^{2a+2} - 4q^{2a+1} + 2q^{2a} - 2q^{a+1} + 1}{q^a (q - 1)(q^{a+1} - 1)},$$ whose partial fraction decomposition, per WolframAlpha, is $$\frac{2(q - 1)}{q(q^{a+1} - 1)} - \frac{1}{q^a (q - 1)} + \frac{3q^2 - 4q + 2}{q(q - 1)},$$ where of course $$\frac{3q^2 - 4q + 2}{q(q - 1)} < I(q^k) + I(n^2)$$ holds unconditionally (i.e. even if we do not have the condition $k \leq a$).
So here is my:
INITIAL QUESTION
Does the inequality $$\frac{2(q - 1)}{q(q^{a+1} - 1)} > \frac{1}{q^a (q - 1)}$$ hold for all special primes $q$?
MY ATTEMPT
Suppose to the contrary that there exists a special prime $q$ such that $$\frac{1}{q^a (q - 1)} \geq \frac{2(q - 1)}{q(q^{a+1} - 1)}.$$
This last inequality is equivalent to $$q^{a+2} - q \geq 2{q^a}(q - 1)^2 = 2{q^a}(q^2 - 2q + 1) = 2q^{a+2} - 4q^{a+1} + 2q^a$$ $$\iff 0 \geq q^{a+2} - 4q^{a+1} + 2q^a + q = q^{a+1} (q - 4) + 2q^a + q > 0,$$ which is a contradiction.
FINAL QUESTION
Does the contradiction suffice to prove that $k \leq a$?
