Informal solution sketch:
Divide into several cases based on the signs of $a, b, c$. But for the all-positive case, it's not so bad. The DC on the left consists of all rationals $x(y)$ with $x \le a$ and $y \le b+c$. Such a rational $y$ can be expressed as $u+v$, with $u, v$ rational, and $u \le b, v \le c$ (because addition on DCs is well-defined). But now look at
x(u+v) = xu + xv
It's a sum of rationals, the first of which is no more than $ab$, and the second of which is more more than $ac$, and hence is a rational that's no more than $ab + ac$, and thus is in the DC representing $ab+ac$. So the DC for the left hand side is a subset of the DC for the right hand side.
Now you have to prove the other direction as well (it's a little easier).
Deep inside this argument is the fact that the distributive law holds for the rationals.
I know this is only a sketch, but it should be enough for you to work out the rest of the details. I've been pretty informal, but the key ideas are there.