Is $D(x) + D(y) \leq D(xy)$ true when $\gcd(x, y)=1$ and $xy$ is deficient?

Question

The title says it all.

Here, $D(z) := 2z - \sigma(z)$ is the deficiency of $z \in \mathbb{N}$, and $\sigma(z)$ is the sum of the divisors of $z$.

Edited title on February 14 2017

ORIGINAL TITLE - Is $D(x) + D(y) \leq D(xy)$ true when $\gcd(x,y)=1$?

Added on February 14 2017

For an infinite family of counterexamples, consider $x=p=2$ and $y=q$ where $p$ and $q$ are distinct primes.

Then $$D(p) + D(q) = (p-1)+(q-1) = q \nleq D(pq) = 2pq - (p+1)(q+1) = (p-1)(q-1) - 2 = q - 3.$$

Revised Question

When does $D(x) + D(y) \leq D(xy)$ hold?

Answer 1

Try the smallest non-trivial example of coprime integers: $$ D(2)+D(3)=1+2\color{red}{\not\le} 0=D(6).$$

Answer 2

Take $2^n$ and $3^n$. Then the inequality do not hold.