 Solution

We first prove the needed Petridis minimal-growth lemma. For finite nonempty sets $A,D$, choose a nonempty $X\subseteq A$ minimizing $K_0=|X+D|/|X|$. Thus $|Y+D|\ge K_0|Y|$ for every $Y\subseteq X$, including the empty set. For any finite $T=\{t_1,\dots ,t_r\}$, defineThe sets $X_i+t_i$ partition $X+T$. If $x\notin X_i$, then $x+t_i\in X+t_j$ for some $j<i$, and therefore $x+D+t_i\subseteq X+D+t_j$. The newly contributed portion of $X+D+t_i$ consequently has size at mostSum over $i$ to obtain $|X+D+T|\le K_0|X+T|$. Apply this successively with $T=(r-1)D$ to get $|X+rD|\le K_0^r|X|$.

Use $D=-A$. The hypothesis implies $K_0\le |A-A|/|A|\le C$, so $|X-2A|\le C^2|X|$. We also prove the Ruzsa triangle inequality: for finite nonempty $U,V,W$,For each $s\in U-W$, fix $u_s,w_s$ with $s=u_s-w_s$. The map $(s,v)\mapsto (u_s-v,v-w_s)$ is injective: summing its coordinates recovers $s$, and then its first coordinate recovers $v$. This proves the inequality.

Apply it with $U=W=2A$ and $V=X$. Both right-hand factors have cardinality $|X-2A|$, givingThus $\boxed{K=C^4\text{ suffices}}$. This is the required case of the Plünnecke-Ruzsa inequality, proved here using the minimal-growth argument and the explicit injection. If $A$ is empty the conclusion holds directly.