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 (sharp?) upper bound $$I(p^k)<\frac{p^{k+1}}{p^k (p - 1)}=\frac{p}{p - 1}.$$ We also have the formula $$\frac{D(p^k)}{p^k}=2-I(p^k),$$ and corresponding (sharp?) upper bound $$\frac{D(p^k)}{p^k}=2-I(p^k) \leq 2-\frac{p+1}{p}=\frac{p-1}{p}.$$
We obtain $$I(p^k)\bigg(2 - I(p^k)\bigg) < \frac{p}{p-1}\cdot\frac{p-1}{p}=1.$$
Here are my questions:
(1) Can one substantially improve on the bound $$I(p^k)\bigg(2 - I(p^k)\bigg) < 1?$$
(2) If the answer to Question (1) is YES, my next question is "How?".
(3) If the answer to Question (1) is NO, can you explain/show why the bound cannot be substantially improved?
MY ATTEMPT
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: $$f(k) = I(p^k)\bigg(2 - I(p^k)\bigg) \leq f(1) = I(p)\bigg(2 - I(p)\bigg) = \frac{p+1}{p}\bigg(\frac{2p - (p+1)}{p}\bigg) = \frac{p^2 - 1}{p^2},$$ which somehow improves on the upper bound of $1$.
