**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 $$I(q^k) + I(n) < I(q^k) + I(n^2) < 3.$$