On bounds for the deficiency of $m^2$, where $p^k m^2$ is an odd perfect number with special prime $p$ Hereinafter, call a number $N$ perfect if $N$ satisfies $\sigma(N)=2N$, where
$$\sigma(x)=\sum_{d \mid x}{d}$$
is the sum of divisors of the positive integer $x$.  Denote the abundancy index of $x$ by $I(x)=\sigma(x)/x$, the deficiency of $x$ by $D(x)=2x-\sigma(x)$, and the sum of aliquot divisors of $x$ by $s(x)=\sigma(x)-x$.
Let $n = p^k m^2$ be an odd perfect number given in Eulerian form, that is, $p$ is the special/Euler prime satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$.
We shall use the following results in deriving bounds for $D(m^2)$:

It turns out that it is possible to express $\gcd(m^2,\sigma(m^2))$ as an integral linear combination of $m^2$ and $\sigma(m^2)$, in terms of $p$ alone.
To begin with, write
$$\gcd(m^2,\sigma(m^2))=\frac{\sigma(m^2)}{p^k}=\frac{D(m^2)}{s(p^k)}=\frac{(2m^2 - \sigma(m^2))(p-1)}{p^k - 1}.$$
Now, using the identity
$$\frac{A}{B}=\frac{C}{D}=\frac{A-C}{B-D},$$
where $B \neq 0$, $D \neq 0$, and $B \neq D$, we obtain
$$\gcd(m^2,\sigma(m^2))=\frac{\sigma(m^2) - (2m^2 - \sigma(m^2))(p-1)}{p^k - (p^k - 1)}$$
so that we get
$$\gcd(m^2,\sigma(m^2))=\frac{\sigma(m^2)}{p^k}=\frac{D(m^2)}{s(p^k)}=2m^2 - pD(m^2)=2(1-p)m^2 + p\sigma(m^2).$$

At once, since $s(p^k) \geq 1$, we have the upper bound:
$$D(m^2) \leq 2m^2 - pD(m^2) \implies D(m^2) \leq \frac{m^2}{(p+1)/2}.$$
Equality holds if and only if the Descartes-Frenicle-Sorli Conjecture that $k=1$ holds.
We now attempt to derive a lower bound for $D(m^2)$ (in terms of $p$, $m^2$ and $\sigma(m^2)$), using the result discussed in this recent MSE question:

From the result
$$s(a)s(b) + (a + b) \leq s(ab)$$
which holds when $\gcd(a,b)=1$, $a>1$, and $b>1$, then setting $a=p^k$ and $b=m^2$, we obtain
$$s(p^k)s(m^2) + (p^k + m^2) \leq s(p^k m^2) = p^k m^2$$
$$\implies 1 + s(p^k)s(m^2) \leq (p^k m^2 - (p^k + m^2) + 1) = (p^k - 1)(m^2 - 1) = (p - 1)(m^2 - 1)s(p^k)$$
$$\implies 1 \leq \bigg((p-1)(m^2 - 1) - s(m^2)\bigg)s(p^k)$$
Multiplying both sides by $D(m^2)$ and dividing through by $s(p^k)$, we get
$$\frac{D(m^2)}{s(p^k)} \leq D(m^2)\cdot{\bigg((p-1)(m^2 - 1) - s(m^2)\bigg)}.$$
But we know from a previous calculation that
$$\frac{D(m^2)}{s(p^k)}=2m^2 - pD(m^2)=2(1-p)m^2 + p\sigma(m^2).$$
Hence, we have the lower bound
$$\frac{2(1-p)m^2 + p\sigma(m^2)}{(p-1)(m^2 - 1) - (\sigma(m^2) - m^2)} \leq D(m^2).$$

Summarizing, we have the bounds:

$$\frac{2(1-p)m^2 + p\sigma(m^2)}{(p-1)(m^2 - 1) - (\sigma(m^2) - m^2)} \leq D(m^2) \leq \frac{m^2}{(p+1)/2}.$$

Here are my questions:

(1) Does anybody here have any bright ideas on how to simplify the lower bound for $D(m^2)$?
(2) Are these bounds best-possible?

 A: We can get a better bound.

To get a better bound, we need a better inequality than $\sigma(x)-x\ge 1$.
So, let us find a better inequality on $\sigma(m^2)$.
To find a better lower bound, let us consider $m$ of the form $PQ$ where $P\lt Q$ are distinct primes.
Then, we have
$$\begin{align}\sigma(m^2)&\ge (1+P+P^2)(1+Q+Q^2)
\\\\&=1+P+P^2+Q+Q^2+PQ(P+Q+1)+P^2Q^2
\\\\&\ge 1+2+2^2+3+3^2+m(2+3+1)+m^2
\\\\&=m^2+6m+19\end{align}$$
from which we have
$$m^2-\sigma(m^2)\le -6m-19$$
Using this, we get, similarly as you did, 
$$\begin{align}&s(p^k)s(m^2) -s(p^km^2)\le -m^2+p^k(-6m-19)
\\\\&\implies s(p^k)s(m^2)\le p^km^2-m^2+p^k(-6m-19)
\\\\&\implies s(p^k)s(m^2)\le (p^k-1)(m^2-1)+p^k(-6m-18)-1
\\\\&\implies s(p^k)s(m^2)\le (p-1)(m^2-1)s(p^k)+p^k(-6m-18)-1
\\\\&\implies p(6m+18)+1\le ((p-1)(m^2-1)-s(m^2))s(p^k)\end{align}$$
Multiplying the both sides by $\frac{D(m^2)}{s(p^k)}$ gives
$$\frac{D(m^2)}{s(p^k)}(p(6m+18)+1)\le ((p-1)(m^2-1)-s(m^2))D(m^2)$$
from which we get
$$\frac{(2(1-p)m^2 + p\sigma(m^2))(p(6m+18)+1)}{(p-1)(m^2-1)-(\sigma(m^2)-m^2)}\le D(m^2)$$
A: The upper bound
$$D(m^2) \leq \frac{m^2}{(p+1)/2}$$
holds if and only if
$$2 - I(m^2) = \frac{D(m^2)}{m^2} \leq \frac{2}{p+1},$$
which is true if and only if
$$\frac{2p}{p+1} = 2 - \frac{2}{p+1} \leq I(m^2).$$
But in general, we know that
$$I(m^2) = \frac{2}{I(p^k)} \leq \frac{2}{I(p)} = \frac{2p}{p+1}.$$
This means that we have
$$I(m^2) = \frac{2}{I(p^k)} = \frac{2p}{p+1} = \frac{2}{I(p)},$$
which is true if and only if $k=1$.
Hence we actually have that
$$\bigg(D(m^2) \leq \frac{m^2}{(p+1)/2}\bigg) \iff \bigg(k=1\bigg).$$

In general, as shown in this closely related question, what is true is that
$$\frac{m^2}{(p+1)/2} \leq D(m^2).$$

