QUESTION
If $N=q^k n^2$ is an odd perfect number, which is bigger, $\sigma(n^2)/n^2$ or $\sigma(n)/q^k$?
BACKGROUND
If $\sigma(N)=2N$ (where $\sigma(N)$ is the sum of the divisors of $N$), then $N$ is said to be perfect. (Denote the abundancy index of $x \in \mathbb{N}$ by $I(x)=\sigma(x)/x$.)
Euler proved that every odd perfect number has the form $N=q^k n^2$, where $q$ is prime (called the Euler prime) satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
MY ATTEMPT TO ANSWER THE QUESTION
Let $N=q^k n^2$ be an odd perfect number with Euler prime $q$.
Suppose that $$\frac{\sigma(n^2)}{n^2}=\frac{\sigma(n)}{q^k}.$$
Then $$\frac{\sigma(q^k)}{n}\cdot\frac{\sigma(n)}{q^k}<2=\frac{\sigma(q^k)}{q^k}\cdot\frac{\sigma(n^2)}{n^2}=\frac{\sigma(q^k)}{q^k}\cdot\frac{\sigma(n)}{q^k}$$ so that $q^k < n$.
Additionally, $$\frac{\sigma(n^2)}{n^2}=\frac{\sigma(n)}{q^k}$$ implies that $${q^k}\sigma(n^2) = {n^2}\sigma(n).$$ Therefore, $n^2 \mid {q^k}\sigma(n^2)$ and $q^k \mid {n^2}\sigma(n)$. Since $\gcd(q,n)=\gcd(q^k, n^2)=1$, then $n^2 \mid \sigma(n^2)$ and $q^k \mid \sigma(n)$. This implies that $$\frac{\sigma(n^2)}{n^2}=\frac{\sigma(n)}{q^k}$$ is an integer, contradicting $$1 < \frac{\sigma(n^2)}{n^2} < 2.$$ (In fact, we have the stronger bound $8/5 < \sigma(n^2)/n^2$.)
Consequently, $$\frac{\sigma(n^2)}{n^2} \neq \frac{\sigma(n)}{q^k}.$$
Therefore, we either have $$\frac{\sigma(n)}{q^k} < \frac{\sigma(n^2)}{n^2} < 2$$ or $$\frac{8}{5} < \frac{\sigma(n^2)}{n^2} < \frac{\sigma(n)}{q^k}.$$
The second case implies $q^k < n$ since $$\frac{\sigma(n)}{n} < \frac{\sigma(n^2)}{n^2} < \frac{\sigma(n)}{q^k}.$$
The first case implies $$n < {2^{\frac{\ln(12/13)}{\ln(9/13)}}}\cdot{(q^k)^{(\frac{\ln(13/12)}{\ln(13/9)}+\ln(13/9))}} \approx (1.162854)\cdot(q^k)^{0.5853949291}$$ if we use the estimate $$\frac{\sigma(n^2)}{n^2} < \bigg(\frac{\sigma(n)}{n}\bigg)^{\frac{\ln(13/9)}{\ln(4/3)}}.$$ Note that the inequality $$n < {2^{\frac{\ln(12/13)}{\ln(9/13)}}}\cdot{(q^k)^{(\frac{\ln(13/12)}{\ln(13/9)}+\ln(13/9))}} \approx (1.162854)\cdot(q^k)^{0.5853949291}$$ almost violates the inequality $q^k < n^2$.
References