A number-theory question on the deficiency function $2x - \sigma(x)$

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: $$D(xy) - D(x)D(y) = 2xy - \sigma(xy) - (2x - \sigma(x))(2y - \sigma(y))$$ $$=2xy - \sigma(xy) - 4xy + 2y\sigma(x) + 2x\sigma(y) - \sigma(x)\sigma(y)$$ $$=-2xy - 2\sigma(x)\sigma(y) + 2y\sigma(x) + 2x\sigma(y) + (\sigma(x)\sigma(y) - \sigma(xy))$$ $$=2(x - \sigma(x))(\sigma(y) - y) + (\sigma(x)\sigma(y) - \sigma(xy)).$$

This is because I want to compare $D(xy)$ and $D(x)D(y)$. Note that, in general we have $$2(x - \sigma(x))(\sigma(y) - y) \leq 0$$ and $$\sigma(x)\sigma(y) - \sigma(xy) \geq 0.$$

Lastly, observe that if we have $D(xy) = D(x)D(y)$, then it follows that $$2(\sigma(x) - x)(\sigma(y) - y) = \sigma(x)\sigma(y) - \sigma(xy).$$

My question is: Will it be possible to prove either $$D(xy) \leq D(x)D(y)$$ or $$D(x)D(y) \leq D(xy),$$ subject to additional assumptions, if necessary?

[Updated Sept. 8, 2015]

If $x$ and $y$ are relatively prime (i.e., $\gcd(x, y) = 1$), then it follows that $D(xy) \leq D(x)D(y)$. By the contrapositive, we have the equivalent implication: "If $D(x)D(y) < D(xy)$, then $\gcd(x, y) > 1$."

Finally, I conjecture that if the equation $D(xy) = D(x)D(y)$ is true, then either $x$ or $y$ is $1$, or both $x$ and $y$ are powers of $2$.

Answer 1

I ran some quick numerics to see what sort of things happen.

You have two questions. Firstly, you conjecture

$D(xy) = D(x)D(y) \implies x$ or $y = 1$, or both $x,y$ are powers of $2$.

I have no intuition for whether this should or shouldn't be true. But the smallest counterexample I found was $15$ and $3$. Note that $D(15) = 30 - (1 + 3 + 5 + 15) = 6$ and $D(3) = 6 - (1 + 3) = 2$. On the other hand, $D(45) = 90 - (1 + 3 + 5 + 9 + 15 + 45) = 12$.

So $D(45) = D(3)D(15)$, and this conjecture is false.

Secondly, you consider the inequalities

Can we say that $D(xy) \geq D(x)D(y)$ or $D(xy) \leq D(x)D(y)$ in a meaningful way?

After the raw numerics, it seems that $D(xy) < D(x)D(y)$ more often. We might expect this, as we know that $$ \sum_{n \leq X} \sigma(n) = \frac{\zeta(2)}{2}X^2 + O(X),$$ leading to the (very loose) heuristic that the average value of $\sigma(n)$ is about $n\zeta(2)/2$. But $D(xy) > D(x)D(y)$ very often, with few discernible patterns (at first glance). It will take (perhaps many or specific) additional constraints to say when either inequality holds.