An improved inequality for the deficiency function when $\gcd(x,y)=1$, $x > 1$, and $y > 1$

Question

(The following is an attempt to improve on the result contained in this MSE question.)

Let $\sigma(x)$ be the sum of the divisors of a (positive) integer $x$. (For example, $\sigma(2) = 1 + 2 = 3$.)

Define the deficiency function $D(x)$ to be the number $$D(x) = 2x - \sigma(x).$$

Let $y$ be a (positive) integer. Now I compute the difference: $$D(xy) - D(x)D(y) = 2xy - \sigma(xy) - (2x - \sigma(x))(2y - \sigma(y))$$ $$=2xy - \sigma(xy) - 4xy + 2y\sigma(x) + 2x\sigma(y) - \sigma(x)\sigma(y)$$ $$=-2xy - 2\sigma(x)\sigma(y) + 2y\sigma(x) + 2x\sigma(y) + (\sigma(x)\sigma(y) - \sigma(xy))$$ $$=2(x - \sigma(x))(\sigma(y) - y) + (\sigma(x)\sigma(y) - \sigma(xy)).$$

Now, assuming that $\gcd(x,y)=1$, the second term vanishes, and we are left with $$D(x)D(y) - D(xy) = 2(\sigma(x) - x)(\sigma(y) - y).$$ If we further assume that $x>1$ and $y>1$ both hold, then we have $$\sigma(x) - x \geq 1$$ $$\sigma(y) - y \geq 1$$ from which it follows that $$D(x)D(y) - D(xy) \geq 2\times{1}\times{1} = 2.$$ We therefore conclude that $$D(x)D(y) \geq D(xy) + 2.$$

Here are my questions:

QUESTIONS

(1) Is the derivation of the improved inequality correct?

(2) Will it be possible to derive a better bound than the improved inequality? Or is this already best-possible?

Answer 1

(1) It looks correct to me.

(2) You cannot get a better inequality than $$D(x)D(y) - D(xy) \ge 2$$ since the equality of this inequality holds when $a,b$ are distinct primes.