Hereinafter, let $I(x)=\sigma(x)/x$ denote the abundancy index of the positive integer $x$, where $\sigma(x)=\sigma_1(x)$ is the classical sum of divisors of $x$. Denote the deficiency of $x$ by
$$D(x)=2x-\sigma(x).$$
Call a positive integer $y$ deficient when $I(y) < 2$. (Equivalently, $y$ is deficient when $D(y) \geq 1$.)
In this paper, the following result is proved:
If $M > 1$ is deficient, then
$$\frac{2M}{M + D(M)} < I(M) < \frac{2M + D(M)}{M + D(M)}.$$
Now, let $N = p^k m^2$ be an odd perfect number given in Eulerian form (i.e. $p$ is the special/Euler prime satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$).
Since $N$ is perfect, and $p^k$ and $m^2$ are proper factors of $N$, then $p^k$ and $m^2$ are deficient. It follows that
$$\frac{2p^k}{p^k + D(p^k)} < I(p^k) < \frac{2p^k + D(p^k)}{p^k + D(p^k)}$$
and
$$\frac{2m^2}{m^2 + D(m^2)} < I(m^2) < \frac{2m^2 + D(m^2)}{m^2 + D(m^2)}.$$
Here is my:
QUESTION: Does the following inequality hold in general, if $p^k m^2$ is an odd perfect number with special/Euler prime $p$?
$$\frac{2p^k + D(p^k)}{p^k + D(p^k)} \leq \frac{2m^2}{m^2 + D(m^2)}$$
MY ATTEMPT
Suppose to the contrary that
$$\frac{2m^2}{m^2 + D(m^2)} < \frac{2p^k + D(p^k)}{p^k + D(p^k)}.$$
This holds if and only if
$$\frac{2}{3 – I(m^2)} = \dfrac{\dfrac{2m^2}{m^2}}{\dfrac{m^2 + D(m^2)}{m^2}} < \dfrac{\dfrac{2p^k + D(p^k)}{p^k}}{\dfrac{p^k + D(p^k)}{p^k}} = \frac{4 – I(p^k)}{3 – I(p^k)}.$$
Since $p^k$ and $m^2$ are deficient, we may cross-multiply, giving
$$\frac{2}{3 – I(m^2)} < \frac{4 – I(p^k)}{3 – I(p^k)} \iff 2(3 – I(p^k)) < (4 – I(p^k))(3 – I(m^2))$$
$$\iff 6 – 2I(p^k) < 12 – 3I(p^k) – 4I(m^2) + I(p^k)I(m^2) = 14 – 3I(p^k) – 4I(m^2)$$
$$\iff 4I(m^2) + I(p^k) < 8.$$
Now, we use the fact that $p$ is the special prime (satisfying $p \equiv 1 \pmod 4$) to derive a lower bound for $p$ (and therefore, an upper bound for $I(p^k)$):
$$p \text{ is prime and } p \equiv 1 \pmod 4 \implies p \geq 5$$
$$I(p^k) = \frac{\sigma(p^k)}{p^k} = \frac{p^{k+1} – 1}{p^k (p – 1)} < \frac{p^{k+1}}{p^k (p – 1)} = \frac{p}{p – 1} \leq \frac{5}{4}.$$
Hence the inequality
$$4I(m^2) + I(p^k) < 8$$
together with the inequality
$$3I(p^k) < \frac{15}{4}$$
implies that
$$4(I(p^k) + I(m^2)) < 8 + \frac{15}{4} = \frac{47}{4}$$
from which we obtain
$$I(p^k) + I(m^2) < \frac{47}{16},$$
which is better than the currently known $I(p^k) + I(m^2) < 3$.
But it is known that
$$3 – \bigg(\frac{p-2}{p(p-1)}\bigg) < I(p^k) + I(m^2).$$
Together with
$$I(p^k) + I(m^2) < \frac{47}{16}$$
this gives
$$3 – \bigg(\frac{p-2}{p(p-1)}\bigg) < \frac{47}{16}$$
$$\iff \frac{1}{16} < \frac{p-2}{p(p-1)}$$
$$\iff p(p – 1) < 16(p – 2)$$
$$\iff p^2 – 17p + 32 < 0$$
$$\iff \frac{17 – \sqrt{161}}{2} < p < \frac{17 + \sqrt{161}}{2}.$$
The lower bound thus obtained for $p$:
$$p > \frac{17 – \sqrt{161}}{2} \approx 2.15571$$
is trivial, and may be replaced by the lower bound $p \geq 5$.
The upper bound thus obtained for $p$:
$$p < \frac{17 + \sqrt{161}}{2} \approx 14.8443$$
is nontrivial.
Note that we then have either
$$p \in \{5, 13\}.$$
If $p = 5$, then
$$\frac{57}{20} < I(5^k) + I(m^2) = I(5^k) + \frac{2}{I(5^k)} = \frac{5^{k+1} – 1}{5^k (5 – 1)} + \frac{2 \bigg({5^k} (5 – 1)\bigg)}{5^{k+1} – 1} = \frac{57 \times 5^{2k} – 2 \times 5^{k+1} + 1}{4\times {5^k} (5^{k+1} – 1)} < \frac{47}{16}.$$
WolframAlpha gives the solution
$$k > \dfrac{\log\bigg(\dfrac{7 + \sqrt{161}}{14}\bigg)}{\log(5)} \approx 0.211864,$$
which is trivial.
If $p = 13$, then
$$\frac{57}{20} < I({13}^k) + I(m^2) = I({13}^k) + \frac{2}{I({13}^k)} = \frac{{13}^{k+1} – 1}{{13}^k (13 – 1)} + \frac{2 \bigg({{13}^k} (13 – 1)\bigg)}{{13}^{k+1} – 1} = \frac{457 \times {13}^{2k} – 2 \times {13}^{k+1} + 1}{{12}\times {{13}^k} ({13}^{k+1} – 1)} < \frac{47}{16}.$$
WolframAlpha gives the solution
$$k > \dfrac{\log\bigg(\dfrac{37 + 3\sqrt{161}}{10}\bigg)}{\log(13)} \approx 0.785894,$$
which is trivial.
Alas, this is where I get stuck.
References:
A criterion for deficient numbers using the abundancy index and deficiency functions, Journal for Algebra and Number Theory Academia, Volume 8, Issue 1, February 2018, pp 1 to 9, preprint available online
The abundancy index of divisors of odd perfect numbers – Part III, Notes on Number Theory and Discrete Mathematics, Volume 23, Number 3, October 2017, pp 53 to 59, available online
Best Answer
The inequality $$\frac{2p^k + D(p^k)}{p^k + D(p^k)} \leq \frac{2m^2}{m^2 + D(m^2)}$$ holds if and only if $$p\ge 13$$
Proof :
We see that $$\frac{2p^k + D(p^k)}{p^k + D(p^k)} \leq \frac{2m^2}{m^2 + D(m^2)}\tag1$$ is equivalent to $$ (p-8)p^{2k+1}+8p^{2k}+2p^k(3p-4)+1\ge 0\tag2$$ because $$\begin{align}(1)&\iff 2m^2(p^k+D(p^k))-(2p^k+D(p^k))(m^2+D(m^2))\ge 0 \\\\&\iff 2m^2p^k+2m^2(2p^k-\sigma(p^k))-2p^km^2-2p^k(2m^2-\sigma(m^2)) \\&\qquad \qquad -m^2(2p^k-\sigma(p^k))-(2p^k-\sigma(p^k))(2m^2-\sigma(m^2))\ge 0 \\\\&\iff 4p^k\sigma(m^2)+m^2\sigma(p^k)-8p^km^2\ge 0 \\\\&\iff 4p^k\frac{2p^km^2}{\sigma(p^k)}+m^2\sigma(p^k)-8p^km^2\ge 0 \\\\&\iff 8p^{2k}+\sigma(p^k)^2-8p^k\sigma(p^k)\ge 0 \\\\&\iff 8p^{2k}+\bigg(\frac{p^{k+1}-1}{p-1}\bigg)^2-8p^k\frac{p^{k+1}-1}{p-1}\ge 0 \\\\&\iff 8p^{2k}(p-1)^2+(p^{k+1}-1)^2-8p^k(p^{k+1}-1)(p-1)\ge 0 \\\\&\iff (p-8)p^{2k+1}+8p^{2k}+2p^k(3p-4)+1\ge 0\end{align}$$
Now,
If $p\ge 13$, then we see that $(2)$ holds.
If $p\lt 13$, i.e. $p=5$, then $(2)$ is equivalent to $5^k(7\cdot 5^{k}-22)\le 1$ which is impossible.
In conclusion, we have $$\frac{2p^k + D(p^k)}{p^k + D(p^k)} \leq \frac{2m^2}{m^2 + D(m^2)}\iff p\ge 13$$