= Solution
The sumset form of <Plünnecke's inequality> states: for nonempty finite sets $A,B$ in an <abelian group>, if $|A+B|\le K|A|$, there is a nonempty $X\subseteq A$ such that
$$
\boxed{|X+mB|\le K^m|X|\quad\text{for every integer }m\ge0.}
$$
Here $0B=\{0\}$. The same $X$ can be used for every $m$. This sumset statement follows from an elementary minimal-growth argument.
Choose $X$ minimizing $\kappa=|X+B|/|X|$ among nonempty subsets of $A$. Then $\kappa\le K$ and $|Y+B|\ge\kappa|Y|$ for every $Y\subseteq X$, including the empty set. We prove the <Petridis minimal-growth lemma>
$$
|X+B+C|\le\kappa|X+C|
$$
for each finite set $C$. Enumerate $C=\{c_1,\ldots,c_t\}$. Put $C_i=\{c_1,\ldots,c_i\}$ and define
$$
X_i=\{x\in X:x+c_i\notin X+C_{i-1}\},\qquad Y_i=X\setminus X_i.
$$
The disjoint new contributions to $X+C$ have sizes $|X_i|$, so $|X+C|=\sum_i|X_i|$. For $y\in Y_i$, the point $y+c_i$ was already present in $X+C_{i-1}$, hence $y+B+c_i\subseteq X+B+C_{i-1}$. Consequently the new contribution at stage $i$ to $X+B+C$ has size at most
$$
|X+B|-|Y_i+B|\le\kappa|X|-\kappa|Y_i|=\kappa|X_i|.
$$
Summing proves the lemma. Starting with $C=\{0\}$ and then taking $C=(m-1)B$ gives inductively $|X+mB|\le\kappa^m|X|\le K^m|X|$, proving the stated <Plünnecke inequality>.
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