Can this inequality involving the deficiency and sum of aliquot divisors be improved? In what follows, we let $n > 1$ be a positive integer.  The classical sum of divisors of $n$ is given by $\sigma_1(n)$.  Denote the abundancy index of $n$ by $I(n)=\sigma_1(n)/n$.
Denote the deficiency of $n$ by $D(n)=2n-\sigma_1(n)$, and denote the sum of aliquot divisors of $n$ by $s(n)=\sigma_1(n)-n$.
CLAIM
$$\frac{D(n^2)}{s(n^2)} < \frac{D(n)}{s(n)}$$
PROOF
$$I(n) < I(n^2) \implies 2 - I(n^2) < 2 - I(n) \implies D(n^2) < nD(n) \implies \frac{D(n^2)}{n^2}<\frac{D(n)}{n}$$
$$I(n) < I(n^2) \implies I(n) - 1 < I(n^2) - 1 \implies ns(n) < s(n^2) \implies \frac{s(n)}{n} < \frac{s(n^2)}{n^2}$$
From the last two inequalities, we get
$$\bigg(\frac{D(n^2)}{n^2}<\frac{D(n)}{n}\bigg) \land \bigg(\frac{n^2}{s(n^2)}<\frac{n}{s(n)}\bigg).$$
Multiplying LHS and RHS of the two inequalities, we finally obtain
$$\frac{D(n^2)}{s(n^2)} < \frac{D(n)}{s(n)}.$$
Here are my questions:

Can the inequality in the CLAIM be improved?  If so, how?

 A: I am yet to find a counterexample for the following, I have not spent much time on your problem so will appreciate if this is not considered an improvement on your original inequality, but none the less I hope that it helps in some small way: 
Denoting the Kronecker delta as follows:
$$\delta \left( x,y \right) =\cases{1&$x=y$\cr 0&$x\neq y $\cr}\tag{                             0}$$
up to $n \leq 2 \cdot 10^7$ I have found the following to be satisfied:
$${\frac {D \left( n \right) }{s \left( n \right) }}-{\frac {
D \left( {n}^{2} \right) }{s \left( {n}^{2} \right) }}-\frac{1}{4}
\delta \left( n-2\,\left\lfloor \frac{n}{2}\right\rfloor,1 \right) \lt \frac{3}{4} \tag{1}$$
A: I am posting this answer in the context of odd perfect numbers $p^k m^2$ with special prime $p$ satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$.
Based from a formula in this recent question, we have
$$D(x)D(y)-D(xy)=2s(x)s(y)$$
when $\gcd(x,y)=1$.
In particular, since $p^k m^2$ is perfect (and $\gcd(p,m)=\gcd(p^k,m^2)=1$), we obtain $D(p^k m^2) = 0$, so that
$$D(p^k)D(m^2)=2s(p^k)s(m^2).$$
This last equation is equivalent to
$$\frac{D(m^2)}{s(m^2)}=\frac{2s(p^k)}{D(p^k)}.$$
But using the same formula, since $\gcd(p,m)=\gcd(p^k,m)=1$ we obtain
$$D(p^k)D(m)-D(p^k m)=2s(p^k)s(m).$$
Dividing throughout the last equation by $D(p^k)s(m)$, we get
$$\frac{D(m)}{s(m)}-\frac{D(p^k m)}{D(p^k)s(m)}=\frac{2s(p^k)}{D(p^k)}.$$
Equating the two expressions for
$$\frac{2s(p^k)}{D(p^k)}$$
we derive
$$\frac{D(m^2)}{s(m^2)}=\frac{D(m)}{s(m)}-\frac{D(p^k m)}{D(p^k)s(m)}.$$
A: This answer adds further details to this earlier answer.
As before, let $p^k m^2$ be an odd perfect number with special prime $p$ satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$.
Note that we have the numerical bounds
$$1 < I(p^k) < \frac{5}{4} < \bigg(\dfrac{8}{5}\bigg)^{\dfrac{\ln(4/3)}{\ln(13/9)}} < I(m) < 2.$$
We have obtained the equation
$$\frac{D(m)}{s(m)}-\frac{D(m^2)}{s(m^2)}=\frac{D(p^k m)}{D(p^k)s(m)}$$
from which we get
$$0 < \frac{D(p^k m)}{D(p^k)s(m)}=\frac{2 - I(p^k)I(m)}{(2 - I(p^k))(I(m) - 1)} < \dfrac{2-\bigg(\dfrac{8}{5}\bigg)^{\dfrac{\ln(4/3)}{\ln(13/9)}}}{\dfrac{3}{4}\bigg(\bigg(\dfrac{8}{5}\bigg)^{\dfrac{\ln(4/3)}{\ln(13/9)}} - 1\bigg)} \approx 1.666929067.$$
