This question is an offshoot of this MSE answer.
Let $\sigma(x)$ be the sum of the divisors of the positive integer $x$. (Denote the abundancy index of $x$ by $I(x)=\sigma(x)/x$.)
If $\sigma(M) = 2M$, then $M$ is said to be perfect.
Currently, as of December 2018, there are $51$ known examples of even perfect numbers -- on the other hand, we still do not know whether there are any odd perfect numbers.
Euler derived the general form that an odd perfect number $N$ must take:
$$N = {q^k}{n^2},$$
where $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n) = 1$. We call $q$ the special or Euler prime of $N$.
Descartes, Frenicle and subsequently Sorli conjectured that $k = 1$.
In [Dris, 2012], it was shown that the implications
$$n < q \Longrightarrow k = 1$$ and $$n < q^2 \Longrightarrow k = 1$$
are true.
Now, note that, since $q$ and $\sigma(q) = q + 1$ are consecutive integers, then the following implications are true.
Case 1: $q^k < n < \sigma(q^k) < \sigma(n) \Longrightarrow k > 1$
Case 2: $n < q^k < \sigma(n) < \sigma(q^k) \Longrightarrow k > 1$
The remaining cases to be considered are:
Case 3: $q^k < \sigma(q^k) < n < \sigma(n) \land k \geq 1$
Case 4: $n < \sigma(n) < q^k < \sigma(q^k) \land k \geq 1$
Case 5: $n < q^k \leq \sigma(n) < \sigma(q^k) \land k \geq 1$
Case 6: $n < q^k < \sigma(q^k) \leq \sigma(n) \land k \geq 1$
mathlove (in the hyperlinked MSE answer) is of the opinion that we should separate the cases in the following way:
Case 1 : $q^k < n < \sigma(q^k) < \sigma(n) \Longrightarrow k > 1$
Case 2 : $n < q^k < \sigma(n) < \sigma(q^k) \Longrightarrow k > 1$
Case 3 : $q^k\lt\sigma(q^k)\lt n\lt\sigma(n)\land k \geq 1$
Case 6 : $n\lt q^k\lt\sigma(q^k)\le\sigma(n)\land k \geq 1$
Case 7 : $n\lt \sigma(n)\le q^k\lt\sigma(q^k)\land k \geq 1$
We want to rule out the following scenario:
Case 6 : $n\lt q^k\lt\sigma(q^k)\le\sigma(n)\land k \geq 1$
in order to show that the biconditional $$q^k < n \iff \sigma(q^k) < \sigma(n) \iff \dfrac{\sigma(q^k)}{n} < \dfrac{\sigma(n)}{q^k}$$ holds.
I noticed that $$\sqrt{\Bigg(\dfrac{8}{5}\Bigg)^{\dfrac{\ln(4/3)}{\ln(13/9)}}} < \sqrt{I(n)} < \sqrt{I(q^k n)} < \dfrac{\dfrac{\sigma(q^k)}{n} + \dfrac{\sigma(n)}{q^k}}{2}$$
But $$\dfrac{\dfrac{\sigma(q^k)}{n} + \dfrac{\sigma(n)}{q^k}}{2} < \dfrac{\dfrac{\sigma(q^k)}{n} + \dfrac{\sigma(n)}{q^k}}{I(q^k n)}$$ holds since $q^k n$ is deficient, being a proper factor of the perfect number $q^k n^2$.
However, we can rewrite this as $$\dfrac{\dfrac{\sigma(q^k)}{n} + \dfrac{\sigma(n)}{q^k}}{I(q^k n)} = \dfrac{\dfrac{\sigma(q^k)}{n} + \dfrac{\sigma(n)}{q^k}}{\Bigg(\dfrac{\sigma(q^k)}{n}\Bigg)\cdot\Bigg(\dfrac{\sigma(n)}{q^k}\Bigg)} = \dfrac{q^k}{\sigma(n)} + \dfrac{n}{\sigma(q^k)}.$$
Hence, we have the simultaneous inequalities: $$\sqrt{\Bigg(\dfrac{8}{5}\Bigg)^{\dfrac{\ln(4/3)}{\ln(13/9)}}} < \dfrac{q^k}{\sigma(n)} + \dfrac{n}{\sigma(q^k)}$$ and $$2\sqrt{\Bigg(\dfrac{8}{5}\Bigg)^{\dfrac{\ln(4/3)}{\ln(13/9)}}} < \dfrac{\sigma(q^k)}{n} + \dfrac{\sigma(n)}{q^k}.$$
Under this scenario:
Case 6 : $n\lt q^k\lt\sigma(q^k)\le\sigma(n)\land k \geq 1$
we obtain the upper bound $$\dfrac{\sigma(q^k)}{n} + \dfrac{\sigma(n)}{q^k} < \dfrac{2}{\sqrt{\Bigg(\dfrac{8}{5}\Bigg)^{\dfrac{\ln(4/3)}{\ln(13/9)}}} - 1} < 10,$$ as $$\dfrac{2}{\sqrt{\Bigg(\dfrac{8}{5}\Bigg)^{\dfrac{\ln(4/3)}{\ln(13/9)}}} - 1} \approx 9.909120785838094255.$$
Here are my:
QUESTIONS Using the ideas in this post, would it be possible to show that $$\dfrac{\sigma(q^k)}{n} + \dfrac{\sigma(n)}{q^k} < 10$$ holds in all cases? If it is not possible, can you explain why?