This post is an offshoot of this MSE question.
Motivation
Let $x, y$ and $z$ be positive integers.
Denote the sum of divisors of $x$ by $\sigma(x)$. Also, denote the deficiency of $y$ by $D(y)=2y-\sigma(y)$. Lastly, denote the abundancy index of $z$ by $I(z)=\sigma(z)/z$.
Let $N = p^k m^2$ be an odd perfect number with special/Euler prime $p$ satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$. Since $p^k$ and $m^2$ are proper factors of the perfect number $N = p^k m^2$, then $p^k$ and $m^2$ are deficient.
From the referenced paper, and using the facts that $D(p^k) > 1$ and $D(m^2) > 1$, we have the bounds $$\frac{2p^k}{p^k + D(p^k)} < I(p^k) < \frac{2p^k + D(p^k)}{p^k + D(p^k)}$$ and $$\frac{2m^2}{m^2 + D(m^2)} < I(m^2) < \frac{2m^2 + D(m^2)}{m^2 + D(m^2)}$$ from which we obtain $$\frac{4p^k m^2}{(p^k + D(p^k))(m^2 + D(m^2))} < I(p^k)I(m^2) = 2 < \frac{(2p^k + D(p^k))(2m^2 + D(m^2))}{(p^k + D(p^k))(m^2 + D(m^2))}.$$
This answer by mathlove improves these bounds to
$$\frac{2n}{n+D(n)}<\frac{(2a+2)n−aD(n)}{(a+1)n+D(n)}<I(n) \tag{1}$$
$$I(n)\lt\frac{(2b+4)n-bD(n)}{(b+2)n+D(n)}\lt \frac{2n+D(n)}{n+D(n)}\tag{2},$$
(where it is implicit that $D(n)>1$) and also shows that it holds for any $a>0,b\gt -1$.
So initially, I have the inequalities $$\dfrac{(2c + 2)p^k - cD(p^k)}{(c + 1)p^k + D(p^k)} < I(p^k) < \dfrac{(2d + 4)p^k - dD(p^k)}{(d + 2)p^k + D(p^k)}$$ $$\dfrac{(2e + 2)m^2 - eD(m^2)}{(e + 1)m^2 + D(m^2)} < I(m^2) < \dfrac{(2f + 4)m^2 - fD(m^2)}{(f + 2)m^2 + D(m^2)}.$$
Per mathlove's result, these inequalities hold for any $c > 0$, $e > 0$, $d > -1$, and $f > -1$. In particular, the inequalities hold for $c = d = e = f = 1$.
We then obtain $$\dfrac{(4p^k - D(p^k))(4m^2 - D(m^2))}{(2p^k + D(p^k)(2m^2 + D(m^2))} < I(p^k)I(m^2) = 2 < \dfrac{(6p^k - D(p^k))(6m^2 - D(m^2))}{(3p^k + D(p^k))(3m^2 + D(m^2))}. \tag3$$
Let us focus on the LHS of Inequality (3) first. We get $$(4p^k - D(p^k))(4m^2 - D(m^2)) < 2(2p^k + D(p^k)(2m^2 + D(m^2))$$ $$4p^k m^2 + 2p^k \sigma(m^2) + 2m^2 \sigma(p^k) + \sigma(p^k)\sigma(m^2) = (2p^k + \sigma(p^k))(2m^2 + \sigma(m^2)) < 2(4p^k - \sigma(p^k))(4m^2 - \sigma(m^2)) = 32p^k m^2 - 8p^k \sigma(m^2) - 8m^2 \sigma(p^k) + 2\sigma(p^k)\sigma(m^2),$$ an inequality which is equivalent to $$30 p^k m^2 = 28p^k m^2 + \sigma(p^k)\sigma(m^2) > 10 p^k \sigma(m^2) + 10 m^2 \sigma(p^k),$$ whence by dividing throughout by $10 p^k m^2$, we obtain the final estimate $$I(m^2) + I(p^k) < 3,$$ which agrees with (and does not improve on) known computations.
Now, let us focus on the RHS of Inequality (3). We derive $$2(5p^k - \sigma(p^k))(5m^2 - \sigma(m^2)) = 2(3p^k + D(p^k))(3m^2 + D(m^2)) < (6p^k - D(p^k))(6m^2 - D(m^2)) = (4p^k + \sigma(p^k))(4m^2 + \sigma(m^2))$$ $$2(25p^k m^2 - 5p^k \sigma(m^2) - 5m^2 \sigma(p^k) + \sigma(p^k)\sigma(m^2)) < 16p^k m^2 + 4p^k \sigma(m^2) + 4m^2 \sigma(p^k) + \sigma(p^k)\sigma(m^2)$$ $$36p^k m^2 = 34p^k m^2 + \sigma(p^k)\sigma(m^2) < 14p^k \sigma(m^2) + 14m^2 \sigma(p^k).$$ Dividing throughout the last inequality by $14 p^k m^2$, we obtain $$\dfrac{18}{7} = \dfrac{36}{14} < I(m^2) + I(p^k),$$ a lower bound which does not improve on the previously known $$\dfrac{57}{20} < I(m^2) + I(p^k).$$
Questions
(1) Is there a $4$-tuple $(c,d,e,f)$ such that the inequalities for $I(p^k)$ and $I(m^2)$ (due to mathlove) result to (an) improved lower (and/or upper) bound(s) for $I(p^k) + I(m^2)$?
(2) If the answer to Question (1) is NO, can you explain why?
Reference
Jose Arnaldo Bebita Dris, A Criterion for Deficient Numbers Using the Abundancy Index and Deficiency Functions, arXiv:1308.6767 [math.NT], 2013-2016; Journal for Algebra and Number Theory Academia, Volume 8, Issue 1 (February 2018), 1-9
