[proofplan]
We view the sets $A, A+B, \dots, A+kB$ as the levels of the commutative addition graph generated by $B$. The key input is the Plünnecke selection principle for this graph: if the $j$-step magnification ratio is $D_j$, then there is a non-empty subset of the bottom level whose $k$-step growth is at most $D_j^{k/j}$. Taking the minimum over all non-empty subsets of $A$ gives $D_k \le D_j^{k/j}$, and the desired monotonicity follows by taking $k$-th roots.
[/proofplan]
[step:Build the finite commutative addition graph]
For each integer $m$ with $0 \le m \le k$, define the finite level
\begin{align*}
X_m := A + mB = \{a + b_1 + \cdots + b_m : a \in A,\ b_1,\dots,b_m \in B\},
\end{align*}
with the convention $X_0 := A$. Define the directed graph $\Gamma$ with vertex set $X_0 \cup X_1 \cup \cdots \cup X_k$ and with a directed edge from $x \in X_m$ to $x+b \in X_{m+1}$ for every $0 \le m < k$ and every $b \in B$. Since $A$ and $B$ are finite, every level $X_m$ is finite. Since $G$ is abelian, the edge maps obtained by adding elements of $B$ commute, so $\Gamma$ is the commutative addition graph generated by $B$.
For a non-empty subset $Z \subset A$ and an integer $m$ with $1 \le m \le k$, the set of vertices reachable from $Z$ in exactly $m$ steps is precisely $Z + mB$. Hence the $m$-th magnification ratio of this graph is
\begin{align*}
D_m = \min_{\varnothing \neq Z \subset A} \frac{|Z + mB|}{|Z|}.
\end{align*}
[/step]
[step:Apply Plünnecke's selection principle at the $j$-th level]
We use the Plünnecke selection principle for finite commutative graphs: if $\Gamma$ is a finite commutative graph with bottom level $X_0$ and $j$-th magnification ratio $D_j$, then for every integer $k \ge j$ there exists a non-empty subset $Y \subset X_0$ such that
\begin{align*}
|\operatorname{Im}^{k}(Y)| \le D_j^{k/j}|Y|,
\end{align*}
where $\operatorname{Im}^{k}(Y)$ denotes the set of vertices reachable from $Y$ by directed paths of length $k$.
The hypotheses of the selection principle are satisfied: the graph is finite by the finiteness of $A$ and $B$, and it is commutative because all edge labels act by translations in the abelian group $G$. Applying the principle to the graph built above gives a non-empty set $Y \subset A$ such that
\begin{align*}
|Y + kB| = |\operatorname{Im}^{k}(Y)| \le D_j^{k/j}|Y|.
\end{align*}
[guided]
The purpose of introducing the addition graph is that Plünnecke's theorem is a statement about commutative layered graphs, while the theorem is phrased in additive notation. The bottom level is $X_0=A$, and moving one level upward means adding one element of $B$.
We invoke the Plünnecke selection principle for finite commutative graphs. It says that if the best $j$-step expansion ratio from the bottom level is $D_j$, then one can choose a non-empty bottom-level subset whose $k$-step expansion is bounded by the power $D_j^{k/j}$. The required hypotheses are exactly the ones established in the previous step: the graph is finite because $A$ and $B$ are finite, and the graph is commutative because translations by elements of $B$ commute in the abelian group $G$.
Thus there exists a non-empty subset $Y \subset A$ satisfying
\begin{align*}
|\operatorname{Im}^{k}(Y)| \le D_j^{k/j}|Y|.
\end{align*}
In the addition graph, a length-$k$ path from $y \in Y$ adds $k$ elements of $B$, so
\begin{align*}
\operatorname{Im}^{k}(Y) = Y + kB.
\end{align*}
Therefore
\begin{align*}
|Y + kB| \le D_j^{k/j}|Y|.
\end{align*}
[/guided]
[/step]
[step:Minimize over subsets to compare $D_k$ and $D_j$]
By definition of $D_k$ as a minimum over all non-empty subsets of $A$, and since $Y \subset A$ is non-empty, we have
\begin{align*}
D_k \le \frac{|Y+kB|}{|Y|}.
\end{align*}
Using the estimate obtained from Plünnecke's selection principle gives
\begin{align*}
D_k \le \frac{|Y+kB|}{|Y|} \le D_j^{k/j}.
\end{align*}
Taking $k$-th roots, which preserves the inequality because both sides are non-negative, yields
\begin{align*}
D_k^{1/k} \le D_j^{1/j}.
\end{align*}
Equivalently,
\begin{align*}
D_j^{1/j} \ge D_k^{1/k}.
\end{align*}
[guided]
The selected set $Y$ is one candidate in the minimization defining $D_k$. Since $D_k$ is the minimum of $|Z+kB|/|Z|$ over all non-empty $Z \subset A$, substituting this particular admissible set $Y$ gives
\begin{align*}
D_k \le \frac{|Y+kB|}{|Y|}.
\end{align*}
The previous step gives the sharper upper bound
\begin{align*}
\frac{|Y+kB|}{|Y|} \le D_j^{k/j}.
\end{align*}
Combining the two inequalities gives
\begin{align*}
D_k \le D_j^{k/j}.
\end{align*}
Both $D_k$ and $D_j$ are non-negative because they are ratios of cardinalities of finite sets with non-empty denominator. Therefore taking $k$-th roots preserves the inequality, and we obtain
\begin{align*}
D_k^{1/k} \le D_j^{1/j}.
\end{align*}
This is exactly the claimed monotonicity, written as
\begin{align*}
D_j^{1/j} \ge D_k^{1/k}.
\end{align*}
[/guided]
[/step]
[step:Specialize to $j=1$]
Taking $j=1$ in the inequality just proved gives
\begin{align*}
D_k^{1/k} \le D_1.
\end{align*}
Raising both sides to the $k$-th power gives
\begin{align*}
D_k \le D_1^k.
\end{align*}
This is the stated particular case.
[/step]
