(Note: The contents of this blog post were taken verbatim from this MSE question, dated January 9, 2019.)
Let $N$ be an odd (positive) integer. If $\sigma(N)=2N$ where $\sigma(N)$ is the sum of the divisors of $N$, then $N$ is called an odd perfect number. Let $I(N)=\sigma(N)/N$ denote the abundancy index of $N$. Likewise, denote the deficiency of $x \in \mathbb{N}$ by $D(x)=2x-\sigma(x)$, and the sum of the aliquot divisors of $x$ by $s(x)=\sigma(x)-x$.
Euler showed that an odd perfect number, if one exists, must have the so-called Eulerian form $N = q^k n^2$, where $q$ is the special/Euler prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
The question on the existence of an odd perfect number is one of the long-standing open problems of number theory.
Since $\gcd(q,n)=1$ and the abundancy index function $I$ is multiplicative, we have
$$\sigma(N)=2N \iff I(N)=2=I(q^k)I(n^2)$$
$$\iff \frac{\sigma(q^k)}{q^k} = I(q^k) = \frac{2}{I(n^2)} = \frac{2n^2}{\sigma(n^2)}.$$
Using the formula
$$\frac{A}{B}=\frac{C}{D}=\frac{C-A}{D-B}$$
and the substitutions $A=\sigma(q^k)$, $B=q^k$, $C=2n^2$, and $D=\sigma(n^2)$, and noting that $C-A > 0$ and $D-B > 0$ (see this paper by Dris (2012) for more information), we obtain
$$\frac{\sigma(q^k)}{q^k} = I(q^k) = \frac{2}{I(n^2)} = \frac{2n^2}{\sigma(n^2)} = \frac{2n^2 - \sigma(q^k)}{\sigma(n^2) - q^k}.$$
Using the known bounds
$$\frac{q+1}{q} \leq I(q^k) < \frac{q}{q-1}$$
we get the system of inequalities
$$(q+1)(\sigma(n^2) - q^k) \leq 2qn^2 - q\sigma(q^k)$$
and
$$(q-1)(2n^2 - \sigma(q^k)) < q(\sigma(n^2) - q^k)$$
from which we get
$$q\sigma(q^{k-2})= q(\sigma(q^{k-1}) - q^{k-1}) = q(s(q^k) - q^{k-1}) = qs(q^k) - q^k \leq qD(n^2) - \sigma(n^2)$$
and
$$qD(n^2) - 2n^2 < qs(q^k) - \sigma(q^k) = q\sigma(q^{k-1}) - \sigma(q^k) = -1,$$
of which the last inequality implies that
$$q < \frac{2n^2 - 1}{D(n^2)} = \frac{2n^2 - 1}{2n^2 - \sigma(n^2)}.$$
The two resulting inequalities may be summarized as follows:
If $N=q^k n^2$ is an odd perfect number given in Eulerian form, then
$$q\sigma(q^{k-2}) + \sigma(n^2) \leq qD(n^2) < 2n^2 - 1.$$
Consequently, we have
$$q\sigma(q^{k-2}) + \sigma(n^2) < 2n^2 - 1 \Longrightarrow D(n^2) > q\sigma(q^{k-2}) + 1.$$
However, this last inequality appears to be trivial, as it is known that
$$\frac{\sigma(n^2)}{q^k}=\frac{2n^2}{\sigma(q^k)}=\gcd(n^2,\sigma(n^2))=\frac{D(n^2)}{s(q^k)}=\frac{2s(n^2)}{D(q^k)}$$
so that, if we are willing to assume the Descartes-Frenicle-Sorli Conjecture that $k=1$, and knowing that this conjecture implies $\sigma(q^k) < n$ (see this paper by Dris (2017) for more information), then we obtain
$$\frac{\sigma(n^2)}{q^k}=\gcd(n^2,\sigma(n^2))=\frac{D(n^2)}{s(q^k)}=\frac{2s(n^2)}{D(q^k)}=\frac{2n^2}{\sigma(q^k)}>2n>\sigma(n).$$
(Please see OEIS sequence A322154 for more information.)
Here is my question:
Is it possible to tweak the argument presented in this MSE post in order to come up with a lower bound better than $D(n^2) > 2n$?
I am guessing our best shot would have to emanate from the inequality
$$\sigma(q^{k-2}) + \frac{\sigma(n^2)}{q} \leq D(n^2)$$
where equality holds if and only if $k=1$ (whence $\sigma(q^{k-2})=0$ since it would be an empty sum).
So suppose that $k>1$. It follows from our method that
$$\sigma(q^{k-2}) + \frac{\sigma(n^2)}{q} < D(n^2) = s(q^k)\frac{\sigma(n^2)}{q^k} = \sigma(q^{k-1})\frac{\sigma(n^2)}{q^k}$$
from which we obtain
$$\frac{\sigma(q^{k-2})}{\sigma(q^{k-1})} + \frac{\sigma(n^2)}{q\sigma(q^{k-1})} < \frac{\sigma(n^2)}{q^k}$$
which is trivial as
$$\sigma(q^{k-2}) < \sigma(q^{k-1})$$
while
$$q^k < q\sigma(q^{k-1}) = q(1 + \ldots + q^{k-1}) = q + \ldots + q^k.$$
