Is $D(x) + D(y) \leq D(xy)$ true when $\gcd(x, y)=1$ and $xy$ is deficient? 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?
 A: Try the smallest non-trivial example of coprime integers:
$$ D(2)+D(3)=1+2\color{red}{\not\le} 0=D(6).$$
A: Take $2^n$ and $3^n$. Then the inequality do not hold.
A: 
When does $D(x) + D(y) \leq D(xy)$ hold?

This answer shows a few necessary conditions.
Claim 1 : If $D(x) + D(y) \leq D(xy),x\gt 1,y\gt 1$ and $\gcd(x,y)=1$, then $x,y$ and $xy$ are deficient.
Proof : We have 
$$\begin{align}&D(x) + D(y) \leq D(xy)
\\\\&\iff 2x-\sigma(x)+2y-\sigma(y)\le 2xy-\sigma(x)\sigma(y)
\\\\&\iff \sigma(x)\sigma(y)-\sigma(x)-\sigma(y)\le 2xy-2x-2y
\\\\&\iff \sigma(x)\sigma(y)-\sigma(x)-\sigma(y)+1\le 2xy-2x-2y+2-1
\\\\&\iff (\sigma(x)-1)(\sigma(y)-1)\le 2(x-1)(y-1)-1
\\\\&\iff \frac{\sigma(x)-1}{x-1}\cdot \frac{\sigma(y)-1}{y-1}\le 2-\frac{1}{(x-1)(y-1)}\ (\lt 2)\end{align}$$
So, since we have
$$\frac{\sigma(x)-1}{x-1}\gt 1\qquad\text{and}\qquad \frac{\sigma(y)-1}{y-1}\gt 1$$
we have to have
$$\frac{\sigma(x)-1}{x-1}\lt 2\qquad\text{and}\qquad \frac{\sigma(y)-1}{y-1}\lt 2,$$
i.e.
$$\sigma(x)\lt 2x-1\lt 2x\qquad\text{and}\qquad \sigma(y)\lt 2y-1\lt 2x$$
Also, we have
$$\begin{align}D(x)+D(y)\le D(xy)&\implies \sigma(xy)\le 2xy-2x-2y+\sigma(x)+\sigma(y)
\\\\&\implies \sigma(xy)\lt 2xy-2x-2y+2x+2y
\\\\&\implies \sigma(xy)\lt 2xy\qquad\square\end{align}$$
Claim 2 : There are infinitely many pairs $(x,y)$ such that $D(x) + D(y) \leq D(xy)$ and $\gcd(x,y)=1$.
Take $x=p,y=q$ where $p,q$ are two distinct odd primes.
$$D(x) + D(y) \leq D(xy)\iff (p-2)(q-2)\ge 3$$
which indeed holds.$\qquad\square$
