In what follows, 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 $x$.
The following is an attempt to improve on the upper bound in this MSE question, for the case of odd perfect numbers.
Per the accepted answer:
$$1 \leq \frac{I(x^2)}{I(x)} \leq \prod_{q}{\frac{q^2 + q + 1}{q^2 + q}} = \frac{\zeta(2)}{\zeta(3)} \approx 1.3684327776\ldots$$
Now, consider an odd perfect number $N = p^k m^2$ with special prime $p$.
It is known that $$I(m) > \left(I(m^2)\right)^{\ln(4/3)/\ln(13/9)} > \left(\frac{2(p-1)}{p}\right)^{\ln(4/3)/\ln(13/9)}$$ and $$I(m^2) = \frac{2}{I(p^k)} \leq \frac{2p}{p+1}.$$
It follows that $$\frac{I(m^2)}{I(m)} < \left(\frac{2p}{p+1}\right)\cdot\left(\frac{p}{2(p-1)}\right)^{\ln(4/3)/\ln(13/9)}.$$
Let $$U_1(p) = \left(\frac{2p}{p+1}\right)\cdot\left(\frac{p}{2(p-1)}\right)^{\ln(4/3)/\ln(13/9)}.$$ Since the derivative $U_1'(p)$ satisfy $$U_1'(p) > 0 \text{ for } p > \frac{\log(52/27)}{\log(13/12)} \approx 8.18821 \text{, in other words, for } p \geq 13$$ and $$U_1'(p) < 0 \text{ for } p < \frac{\log(52/27)}{\log(13/12)} \text{, in other words, for } p = 5$$ then because $$1.15387720527781077810718331 \approx U_1(5) < \lim_{p \rightarrow \infty}{U_1(p)} = 2^{\log(13/12)/\log(13/9)} \approx 1.16285$$ we finally obtain $$\frac{I(m^2)}{I(m)} < 2^{\log(13/12)/\log(13/9)}.$$
Note that, if $k=1$, we get $$I(m^2) = \frac{2p}{p + 1}$$ so that $$\frac{I(m^2)}{I(m)} \leq \left(\frac{2p}{p+1}\right)\left(\frac{p+1}{2p}\right)^{\ln(4/3)/\ln(13/9)} = U_2(p).$$
Since the derivative $U_2'(p)$ satisfies $U_2'(p) > 0$ for $p \geq 5$, then $$\frac{I(m^2)}{I(m)} < \lim_{p \rightarrow \infty}{U_2(p)} = 2^{\log(13/12)/\log(13/9)}.$$
Here is my inquiry:
Do you see a(nother) way to improve on the bound $$\frac{I(m^2)}{I(m)} < 2^{\log(13/12)/\log(13/9)}?$$
