The topic of odd perfect numbers likely needs no introduction.
Here is the:
PROBLEM
If $q^k n^2$ is an odd perfect number with special prime $q$, does $q^k < n$ imply that $\sigma(q^k) < n$?
$\sigma(x)$ is the sum of divisors of the positive integer $x$. The special prime $q$ satisfies $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
From the divisibility constraint $\gcd(q,n)=1$, we immediately get $q^k \neq n$. Suppose now that we have $q^k < n$. Is it then possible to obtain $\sigma(q^k) < n$?
MY ATTEMPT
Assume that $q^k < n$. This is equivalent to $$\frac{1}{n} < \frac{1}{q^k}$$ which in turn is equivalent to $$\frac{\sigma(q^k)}{n} < \frac{\sigma(q^k)}{q^k}.$$ But from the formula for the divisor sum $\sigma$, we get (since $q$ is prime) $$\frac{\sigma(q^k)}{q^k} = \frac{q^{k+1} - 1}{q^k (q - 1)} < \frac{q^{k+1}}{q^k (q - 1)} = \frac{q}{q - 1} = \frac{1}{1 - \frac{1}{q}}.$$ Since $q$ is a prime satisfying $q \equiv 1 \pmod 4$, then we have $q \geq 5$, which is equivalent to $$\frac{1}{q} \leq \frac{1}{5} \iff 1 - \frac{1}{q} \geq \frac{4}{5} \iff \frac{1}{1 - \frac{1}{q}} \leq \frac{5}{4}.$$
Hence, we conclude that:
If $q^k < n$, then $\sigma(q^k) < \dfrac{5n}{4}$.
QUESTIONS
(1) Will it be possible to tweak this argument to produce an unconditional proof for $\sigma(q^k) < n$?
(2) If an unconditional proof for $\sigma(q^k) < n$ is not possible, under what conditions can it be proved?
This is a partial answer.
Under the assumption $q^k < n$, it is possible to prove $\sigma(q^k) < n$ when $k=1$.
To see why, assume that $q = q^k < n$. Assume to the contrary that $n < \sigma(q^k) = \sigma(q) = q+1$. Together, the two inequalities yield $$q < n < q+1,$$ contradicting the fact that $n$ is an integer. This concludes the proof.