= Solution
Apply <Plünnecke's inequality> with the summand set also equal to $A$. There is a nonempty $X\subseteq A$ satisfying $|X+mA|\le C^m|X|$ for every $m\ge0$.
We need the <Ruzsa triangle inequality> in a form whose short proof also fixes the signs. For every $z\in R-S$, choose one representation $z=r_z-s_z$. The map
$$
(z,x)\longmapsto(r_z+x,s_z+x),\qquad x\in X,
$$
is injective into $(R+X)\times(S+X)$: its difference recovers $z$, then the fixed representation recovers $x$. Therefore
$$
|R-S||X|\le|R+X|\,|S+X|.
$$
Use $R=rA$ and $S=sA$ and the common set $X$ from the previous part:
$$
|rA-sA|\le\frac{|rA+X|\,|sA+X|}{|X|}
\le C^{r+s}|X|\le C^{r+s}|A|.
$$
Thus
$$
\boxed{|rA-sA|\le C^{r+s}|A|.}
$$
The shared minimizing subset is important; applying separate growth estimates with unrelated subsets would not justify this bound.
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