If $q^k n^2$ is an odd perfect number with Euler prime $q$, is $\sigma(q^k)/n + \sigma(n)/q^k$ bounded from above?

Question

STATEMENT OF THE PROBLEM

If $q^k n^2$ is an odd perfect number with Euler prime $q$, is $\sigma(q^k)/n + \sigma(n)/q^k$ bounded from above?

MOTIVATION

Let $\sigma=\sigma_{1}$ denote the classical sum-of-divisors function, and denote the abundancy index of $x \in \mathbb{N}$ by $I(x)=\sigma(x)/x$.

It is known that the inequality $$I(q^k) + I(n) < \frac{\sigma(q^k)}{n}+\frac{\sigma(n)}{q^k}$$ holds if and only if the biconditional $$q^k < n \iff \sigma(q^k) < \sigma(n)$$ is true. This biconditional is true if $\sigma(q^k)<n$, or if $\sigma(n) \leq q^k$. (I currently am not aware of any other conditions for which the biconditional holds.) Edit (August 11 2017): The biconditional is also true when $q^k < n$. (This follows from $I(q^k)<I(n)$.)

Note that if $\sigma(q^k)/n + \sigma(n)/q^k < C$ for some absolute constant $C$, then $$\sqrt{\frac{8}{5}}\frac{n}{C} < q^k < Cn,$$ so that $C > \sqrt[4]{8/5}$.

However, I know that $C > \sqrt[4]{8/5}$ is far from the truth, as I have recently been able to verify that either $$\frac{\sigma(q^k)}{n} < \sqrt{2} < \frac{\sigma(n)}{q^k}$$ or $$\frac{\sigma(n)}{q^k} < \sqrt{2} < \frac{\sigma(q^k)}{n}$$ is true. In the first case, $q^k < n\sqrt{2}$, while in the second case, we have $n < q^k$.

Of course, trivially we have $$I(q^k) + I(n) < I(q^k) + I(n^2) < 3.$$

Answer 1

Suppose to the contrary that $$\frac{\sigma(q^k)}{n}+\frac{\sigma(n)}{q^k}$$ is bounded from above.

Since $x < \sigma(x)$ for all $x > 1$, and $\sigma(y) < 2y$ for deficient $y$, $$\frac{q^k}{n}+\frac{n}{q^k}<\frac{\sigma(q^k)}{n}+\frac{\sigma(n)}{q^k}<2\cdot\bigg(\frac{q^k}{n}+\frac{n}{q^k}\bigg),$$ so that the biconditional $$\frac{\sigma(q^k)}{n}+\frac{\sigma(n)}{q^k} \text{ is bounded from above } \iff \frac{q^k}{n}+\frac{n}{q^k} \text{ is bounded from above }$$ holds.

But in general, we know that the function $$f(z) = z + \frac{1}{z}$$ is not bounded from above. (It suffices to let $z$ approach $0$ [from the right] and to let $z$ approach $\infty$.)

Consequently, the sum $$\frac{\sigma(q^k)}{n}+\frac{\sigma(n)}{q^k}$$ may not be bounded from above.