Is it possible to improve the resulting upper bound for $\frac{D(m)}{s(m)}$, given a lower bound for $I(m)$? Let $\sigma=\sigma_{1}$ be the classical sum of divisors.  For example,
$$\sigma(12)=1+2+3+4+6+12=28.$$
Define the following arithmetic functions:
$$D(n)=2n-\sigma(n)$$
$$s(n)=\sigma(n)-n$$
$$I(n)=\dfrac{\sigma(n)}{n}.$$
Here is my initial question:
QUESTION

Is it possible to improve the resulting upper bound for $\dfrac{D(m)}{s(m)}$, given a lower bound for $I(m)$?

MY ATTEMPT
For example, assume that a lower bound for $I(m)$ is given as
$$I(m) > c$$
where $1 < c \in \mathbb{R}$.
We rewrite
$$\dfrac{D(m)}{s(m)}$$
as
$$\dfrac{D(m)}{s(m)}=\dfrac{2m-\sigma(m)}{\sigma(m)-m}=\dfrac{2-I(m)}{I(m)-1} < \frac{2-c}{c-1},$$
since
$$\bigg(I(m) - 1 > c - 1\bigg) \land \bigg(2 - I(m) < 2 - c\bigg) \iff \dfrac{2-I(m)}{I(m)-1} < \frac{2-c}{c-1}.$$
Here is my follow-up question:

Can we do better than the upper bound
$$\dfrac{D(m)}{s(m)} < \frac{2-c}{c-1},$$
if $I(m) > c$ (where $1 < c \in \mathbb{R}$)?

I am under the impression that one can come up with a tighter bound.

You are looking for $f(c)$ such that $$\frac{D(m)}{s(m)}<f(c)<\frac{2−c}{c−1}$$
This is equivalent to
$$c<1+\frac{1}{f(c)+1}<I(m)$$
since we have
$$\begin{align}&\frac{D(m)}{s(m)}<f(c)<\frac{2−c}{c−1}
\\\\&\iff -1+\frac{1}{I(m)-1}\lt f(c)\lt -1+\frac{1}{c-1}
\\\\&\iff \frac{1}{I(m)-1}\lt f(c)+1\lt \frac{1}{c-1}
\\\\&\iff c-1\lt \frac{1}{f(c)+1}\lt I(m)-1
\\\\&\iff c\lt 1+\frac{1}{f(c)+1}\lt I(m)\end{align}$$
So, you are looking for $g(c)$ such that $c<g(c)<I(m)$ under the condition that $c<I(m)$.
As you commented, the above means that one can obtain an improved upper bound for $\dfrac{D(m)}{s(m)}$ if and only if one can improve the lower bound for $I(m)$.
