Preamble: This post is an offshoot of this earlier question.
The topic of odd perfect numbers likely needs no introduction.
Denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$, the abundancy index of $x$ by $I(x)=\sigma(x)/x$, and the deficiency of $x$ by $D(x)=2x-\sigma(x)$.
Euler proved that a hypothetical odd perfect number must take the form $$n = p^k m^2$$ where the Eulerian component $p^k$ satisfies the constraints $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$.
Since $p$ is (the special) prime, we have the formula $$I(p^k)=\frac{\sigma(p^k)}{p^k}=\frac{p^{k+1} - 1}{p^k (p - 1)}$$ and corresponding (improved) upper bound $$I(p^k) < \bigg(\frac{p}{p - 1}\bigg)\bigg(\frac{p^{k+1}}{p^{k+1} + 1}\bigg),$$ per this answer to a closely related question. We also have the formula $$\frac{D(p^k)}{p^k}=2-I(p^k),$$ and corresponding (improved) lower bound $$\frac{D(p^k)}{p^k}=2-I(p^k) > 2-\bigg(\frac{p}{p - 1}\bigg)\bigg(\frac{p^{k+1}}{p^{k+1} + 1}\bigg)=\frac{p^{k+2} - 2p^{k+1} + 2p - 2}{(p - 1)(p^{k+1} + 1)}.$$
As in the earlier question, I am aware of the fact that $$f(k):=g(p):=I(p^k)\bigg(2 - I(p^k)\bigg)=\frac{\sigma(p^k) D(p^k)}{p^{2k}}=\frac{(p^{k+1} - 1)(p^{k+1} - 2p^k + 1)}{\Bigg(p^k (p - 1)\Bigg)^2}$$ and that $$\frac{\partial f}{\partial k} = -\frac{2(p^k - 1)\log(p)}{\Bigg(p^k (p - 1)\Bigg)^2} < 0$$ while $$\frac{\partial g}{\partial p} = \frac{2(p^k - 1)(p^{k+1} - (k+1)p + k)}{p^{2k+1} (p - 1)^3} > 0.$$
This means that $$g(5) \leq g(p) = f(k) \leq f(1)$$ since the computations above show that $f(k)$ is decreasing while $g(p)$ is increasing.
In particular, the quantity $f(1)$ in the inequality $f(k) \leq f(1)$ simplifies to: $$f(k) = I(p^k)\bigg(2 - I(p^k)\bigg) \leq f(1) = I(p)\bigg(2 - I(p)\bigg) = \frac{p+1}{p}\bigg(\frac{2p - (p+1)}{p}\bigg) = \frac{p^2 - 1}{p^2}.$$
But from the considerations above, we have $$\bigg(\frac{p+1}{p}\bigg)\cdot\bigg(\frac{p^{k+2} - 2p^{k+1} + 2p - 2}{(p - 1)(p^{k+1} + 1)}\bigg) < I(p^k)\bigg(2 - I(p^k)\bigg) \leq \frac{p^2 - 1}{p^2}$$ which implies that $$\frac{p^{k+2} - 2p^{k+1} + 2p - 2}{(p - 1)(p^{k+1} + 1)} < \frac{p - 1}{p}$$ $$p^{k+3} - 2p^{k+2} + 2p^2 - 2p < (p - 1)^2 \bigg(p^{k+1} + 1\bigg) = p^{k+3} - 2p^{k+2} + p^{k+1} + p^2 - 2p + 1,$$ simplifying to $$p^{k+1} - p^2 + 1 > 0. \text{ (Inequality A) }$$
Note that Inequality A is unconditionally true (i.e. for $k \geq 1$).
Here is my:
QUESTION: Do you see a way to improve on this proof argument (and therefore, on Inequality A), perhaps using ideas from this closely related question, where I derived ever-increasing lower bounds for $I(m^2)$ and therefore, recursively improved upper bounds for $I(p^k)$, by noting that $$I(p^k) = \frac{2}{I(m^2)}?$$
