If $N = q^k n^2$ is an odd perfect number with Euler prime $q$, and $k=1$, does it follow that $\frac{\sigma(n^2)}{n^2} \geq 2 - \frac{5}{3q}$? Let $\sigma=\sigma_{1}$ be the classical sum-of-divisors function.
If $N = q^k n^2$ is an odd perfect number with Euler prime $q$ (i.e., $q$ satisfies $q \equiv k \equiv 1 \pmod 4$, and $\gcd(q,n)=1$), and $k=1$ holds, does it follow that $$\frac{\sigma(n^2)}{n^2} \geq 2 - \frac{5}{3q}?$$
Note that, since the Euler prime $q$ satisfies $q \equiv 1 \pmod 4$, then
$$q \geq 5 \implies \frac{\sigma(n^2)}{n^2} \geq 2 - \frac{5}{3q} \geq 2 - \frac{1}{3} = \frac{5}{3},$$
which agrees with what is known about $N = q^k n^2$ when $k=1$.
In fact, when $k=1$,
$$\frac{\sigma(q^k)}{q^k} = \frac{\sigma(q)}{q} = \frac{q+1}{q} = 1 + \frac{1}{q} \leq 1 + \frac{1}{5} = \frac{6}{5}.$$
$\sigma(n^2)/n^2 \geq 5/3$ then follows from
$$\left(\frac{\sigma(q^k)}{q^k}\right)\cdot\left(\frac{\sigma(n^2)}{n^2}\right)=\left(\frac{\sigma(q)}{q}\right)\cdot\left(\frac{\sigma(n^2)}{n^2}\right)=2.$$

If $N$ is perfect then $2N = 2q^k n^2 = \sigma(q^k n^2) = \sigma(q^k)\sigma(n^2) = \frac{q^{k+1}-1}{q-1} \sigma(n^2)$, which gives
\begin{equation}
\frac{\sigma(n^2)}{n^2} = \frac{2q^k(q-1)}{q^{k+1}-1} = 2-2\frac{q^k-1}{q^{k+1}-1}
\end{equation}
This being $\ge 2-\frac{5}{3q}$ works out to being equivalent to $q^{k+1}-6q+5 \le 0$, which can't work if $k > 1$ for any prime.
