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