Improving on the lower bound $I(q^k) + I(n^2) > 3 - \frac{q-2}{q(q-1)}$, where $q^k n^2$ is an odd perfect number

(Note:  This post is copied verbatim from this MSE question.)

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

$$I(q^k) + I(n^2) > 3 - \frac{q-2}{q(q-1)} = \frac{3q^2 - 4q + 2}{q(q - 1)}.$$

Now, a recent MO post improves on the lower bound for $I(n^2)$, as follows:

$$I(n^2) > \bigg(\frac{2(q-1)}{q}\bigg)\bigg(\frac{q^{k+1} + 1}{q^{k+1}}\bigg)$$

Repeating the same procedure as above, we have the negative product

$$\Bigg(I(q^k) - \left(\frac{2(q-1)}{q}\bigg)\bigg(\frac{q^{k+1} + 1}{q^{k+1}}\right)\Bigg)\Bigg(I(n^2) - \left(\frac{2(q-1)}{q}\bigg)\bigg(\frac{q^{k+1} + 1}{q^{k+1}}\right)\Bigg) < 0.$$

This implies, after some algebraic manipulations, that

$$I(q^k) + I(n^2) > \frac{q^{k+2}}{(q - 1)(q^{k+1} + 1)} + \frac{2(q-1)(q^{k+1} + 1)}{q^{k+2}}.$$

But WolframAlpha says that the partial fraction decomposition of the new lower bound is given by

$$\frac{q^{k+2}}{(q - 1)(q^{k+1} + 1)} + \frac{2(q-1)(q^{k+1} + 1)}{q^{k+2}} = \frac{3q^2 - 4q + 2}{q(q - 1)} + \frac{2(q - 1)}{q^{k+2}} - \frac{q}{(q - 1)(q^{k+1} + 1)}.$$

So essentially, my question boils down to:

QUESTION: Is it possible to produce an unconditional proof (that is, for all $k \geq 1$ and for all special primes $q \geq 5$) for the following inequality?

$$\frac{2(q - 1)}{q^{k+2}} > \frac{q}{(q - 1)(q^{k+1} + 1)}$$

MY ATTEMPT

I tried to ask WolframAlpha for a plot of the above inequality, it gave me the following GIF image:

WolframAlpha Inequality Plot

So it does appear that the inequality is unconditionally true, which would mean that the new lower bound for $I(q^k) + I(n^2)$ improves on the old.  Is it possible to prove this analytically?

And lastly:  Based on this answer to a closely related question, since we appear to have obtained an improved lower bound for $I(q^k) + I(n^2)$, can we then say that there is indeed an integer $a$ such that $k \leq a$?

Posted Answer

Suppose to the contrary that there exists an integer $k \geq 1$ and a (special) prime $q \geq 5$ such that

$$\frac{q}{(q-1)(q^{k+1}+1)} \geq \frac{2(q-1)}{q^{k+2}}.$$

This inequality is equivalent to

$$q^{k+3} \geq 2(q-1)^2 (q^{k+1}+1) = 2q^{k+1} - 4q^{k+2} + 2q^{k+3} + 2q^2 - 4q + 2,$$

which in turn is equivalent to

$$0 \geq q^{k+3} - 4q^{k+2} + 2q^{k+1} + 2q^2 - 4q + 2 = q^{k+2} (q - 4) + 2q^{k+1} + 2q(q - 2) + 2 > 0,$$

a contradiction.

We therefore conclude that

$$\frac{2(q-1)}{q^{k+2}} > \frac{q}{(q-1)(q^{k+1}+1)}$$

for all integers $k \geq 1$ and all (special) primes $q \geq 5$.

Hence, the new lower bound

$$I(q^k) + I(n^2) > \frac{q^{k+2}}{(q - 1)(q^{k+1} + 1)} + \frac{2(q-1)(q^{k+1} + 1)}{q^{k+2}}$$

does indeed improve on the old lower bound

$$I(q^k) + I(n^2) > 3 - \frac{q-2}{q(q - 1)} = \frac{3q^2 - 4q + 2}{q(q - 1)}.$$

It remains to be seen whether this implies that there does exist an integer $a$ such that $k \leq a$, if $q^k n^2$ is an odd perfect number with special prime $q$.