(Note: This has been cross-posted to MO.)
Suppose I have the following series of implications (the underlying assumption is $I(q^k)I(n^2) = 2$), where $I(x) = \sigma(x)/x$ is the abundancy index of the positive integer $x$, and $\sigma$ is the classical sum-of-divisors function:
$$\frac{2n}{n + 1} < I(n^2) \Longrightarrow n < q \Longrightarrow k = 1 \Longrightarrow \frac{2n}{n + 1}\left(\frac{q + 1}{q}\right) = \frac{2n}{n + 1} \cdot I(q) < I(q)I(n^2)$$
$$\Longrightarrow \frac{2n}{n + 1}\left(\frac{q + 1}{q}\right) = \frac{2n}{n + 1} \cdot I(q) < I(q)I(n^2) = I(q^k)I(n^2) = 2$$
$$\Longrightarrow \frac{2n\left(q + 1\right)}{\left(n + 1\right)q} < 2 \Longrightarrow 2qn + 2n < 2qn + 2q \Longrightarrow n < q.$$
My question is: Does this series of implications mean that the biconditional
$$n < q \Longleftrightarrow k = 1$$
is true?
Thanks!
[Edit (February 13, 2015): As mentioned in the comments, the truth of the reverse implication $k = 1 \Longrightarrow n < q$ would depend on whether the inequality
$$\frac{2n}{n + 1} < I(n^2)$$
is indeed true. Per the answer to this MO question, the limiting proportion of $n$ satisfying this last inequality is $0$. Since this inequality is equivalent to $n < q$, heuristically this means that we can reasonably expect $q < n$ to be true, although this is far from a rigorous proof.]
[Update as of March 05, 2020 (Brown, 2016), (Dris, 2017), and (Starni, 2018) have all claimed/published (independent) proofs for the inequality $q<n$, where $q^k n^2$ is an odd perfect number with special/Euler prime $q$.]