[proofplan] We first prove Ruzsa's triangle inequality by an explicit injection and use it to record the standard small-doubling consequence $|A-A| \leq K^2|A|$. The final estimate then follows from the sum-difference form of the Plünnecke-Ruzsa inequality, applied directly to the finite non-empty set $A$ under the hypothesis $|A+A| \leq K|A|$. Since $K \geq 1$, the sharper Plünnecke-Ruzsa exponent $m+n$ implies the stated exponent $2(m+n)-1$. [/proofplan] [step:Prove Ruzsa's triangle inequality by an injection] Let $X,Y,Z \subset G$ be finite non-empty subsets. We claim that \begin{align*} |X-Z|\,|Y| \leq |X-Y|\,|Y-Z|. \end{align*} For each $d \in X-Z$, choose one pair $(x_d,z_d) \in X \times Z$ such that $d=x_d-z_d$. Define \begin{align*} \Phi : (X-Z) \times Y &\to (X-Y)\times (Y-Z) \\ (d,y) &\mapsto (x_d-y,\ y-z_d). \end{align*} This map is well-defined because $x_d-y \in X-Y$ and $y-z_d \in Y-Z$. We verify that $\Phi$ is injective. Suppose \begin{align*} \Phi(d,y)=\Phi(d',y') \end{align*} for $(d,y),(d',y') \in (X-Z)\times Y$. Then \begin{align*} x_d-y &= x_{d'}-y',\\ y-z_d &= y'-z_{d'}. \end{align*} Adding the two equalities in the abelian group $G$ gives \begin{align*} x_d-z_d=x_{d'}-z_{d'}. \end{align*} Thus $d=d'$. Since the chosen representative pair depends only on $d$, we have $x_d=x_{d'}$ and $z_d=z_{d'}$. The equality $x_d-y=x_d-y'$ then gives $y=y'$. Hence $\Phi$ is injective, so \begin{align*} |(X-Z)\times Y| \leq |(X-Y)\times (Y-Z)|. \end{align*} Taking cardinalities of finite Cartesian products yields \begin{align*} |X-Z|\,|Y| \leq |X-Y|\,|Y-Z|. \end{align*} [/step] [step:Control the first difference set $A-A$] Apply the inequality from the previous step with $X=A$, $Y=-A$, and $Z=A$, where \begin{align*} -A := \{-a : a \in A\}. \end{align*} Since $A$ is finite and non-empty, so is $-A$, and $|-A|=|A|$. We obtain \begin{align*} |A-A|\,|A| \leq |A-(-A)|\,|(-A)-A|. \end{align*} Now \begin{align*} A-(-A)=A+A, \qquad (-A)-A=-(A+A), \end{align*} and negation is a bijection on $G$, so $|(-A)-A|=|A+A|$. Therefore \begin{align*} |A-A| \leq \frac{|A+A|^2}{|A|} \leq \frac{(K|A|)^2}{|A|} =K^2|A|. \end{align*} Also, $K\geq 1$: fixing any $a_0\in A$, the translate $A+a_0$ is contained in $A+A$ and has cardinality $|A|$, so $|A|\leq |A+A|\leq K|A|$. [/step] [step:Apply the Plünnecke-Ruzsa sum-difference inequality and weaken the exponent] We use the [Plünnecke-Ruzsa Inequality](/page/Plünnecke%E2%80%93Ruzsa%20Inequality) in its sum-difference form: if $B$ is a finite non-empty subset of an abelian group and $|B+B|\leq L|B|$, then for all integers $p,q\geq 0$ with $p+q\geq 1$, \begin{align*} |pB-qB|\leq L^{p+q}|B|. \end{align*} Apply this result with $B:=A$, $L:=K$, $p:=m$, and $q:=n$. Its hypotheses are satisfied because $A$ is a finite non-empty subset of the abelian group $G$, and the theorem assumption gives $|A+A|\leq K|A|$. Hence \begin{align*} |mA-nA|\leq K^{m+n}|A|. \end{align*} By the previous step, $K\geq 1$. Since $m+n\geq 1$, we have $m+n\leq 2(m+n)-1$, and therefore \begin{align*} K^{m+n}|A|\leq K^{2(m+n)-1}|A|. \end{align*} Combining the two inequalities gives \begin{align*} |mA-nA|\leq K^{2(m+n)-1}|A|, \end{align*} which is the desired estimate. [/step]