[proofplan] We prove the cardinality inequality by constructing an injective map from $(A-C)\times B$ into $(A-B)\times(B-C)$. For each element of $A-C$, we fix one representation as a difference $a-c$; this chosen representation is what allows recovery of the original pair after applying the map. Counting the domain and codomain of the injection gives the desired inequality, and the logarithmic form follows by dividing by $|A|^{1/2}|C|^{1/2}$ and taking logarithms. [/proofplan] [step:Choose one representation for each element of $A-C$] Since $A$ and $C$ are non-empty, every $x \in A-C$ has at least one representation $x=a-c$ with $a \in A$ and $c \in C$. For each $x \in A-C$, choose one such representation and denote it by \begin{align*} x = a_x-c_x, \end{align*} where $a_x \in A$ and $c_x \in C$. [guided] The set $A-C$ may contain the same element in many different ways: it can happen that $a-c=a'-c'$ with $(a,c)\neq(a',c')$. To avoid ambiguity later, we make one fixed choice for every element of $A-C$. Thus, for each $x \in A-C$, choose a pair $(a_x,c_x) \in A \times C$ such that \begin{align*} x = a_x-c_x. \end{align*} This notation means that once $x$ is known, the selected elements $a_x$ and $c_x$ are also known. The finiteness of $A-C$ ensures that these choices are made over a finite set. [/guided] [/step] [step:Embed $(A-C)\times B$ into $(A-B)\times(B-C)$] Define the map \begin{align*} \Phi : (A-C)\times B &\to (A-B)\times(B-C) \\ (x,b) &\mapsto (a_x-b,\ b-c_x). \end{align*} This map is well-defined: if $x \in A-C$ and $b \in B$, then $a_x \in A$ gives $a_x-b \in A-B$, while $c_x \in C$ gives $b-c_x \in B-C$. We prove that $\Phi$ is injective. Suppose $\Phi(x,b)=\Phi(y,b')$ for $(x,b),(y,b') \in (A-C)\times B$. Then \begin{align*} a_x-b &= a_y-b', \\ b-c_x &= b'-c_y. \end{align*} Adding these two equalities in the abelian group $G$ gives \begin{align*} (a_x-b)+(b-c_x)=(a_y-b')+(b'-c_y). \end{align*} Using associativity and cancellation in $G$, this becomes \begin{align*} a_x-c_x=a_y-c_y. \end{align*} By the defining choices, $a_x-c_x=x$ and $a_y-c_y=y$, hence $x=y$. Since the chosen representation depends only on $x$, we have $a_x=a_y$ and $c_x=c_y$. Substituting $a_x=a_y$ into $a_x-b=a_y-b'$ yields $b=b'$. Therefore $(x,b)=(y,b')$, so $\Phi$ is injective. [guided] The goal is to compare the size of $A-C$ with the sizes of $A-B$ and $B-C$. The auxiliary set $B$ is inserted by considering pairs $(x,b)$ with $x \in A-C$ and $b \in B$. Define \begin{align*} \Phi : (A-C)\times B &\to (A-B)\times(B-C) \\ (x,b) &\mapsto (a_x-b,\ b-c_x). \end{align*} The first coordinate belongs to $A-B$ because $a_x \in A$ and $b \in B$. The second coordinate belongs to $B-C$ because $b \in B$ and $c_x \in C$. Hence $\Phi$ is a well-defined map into $(A-B)\times(B-C)$. We now show that no two different input pairs have the same image. Assume \begin{align*} \Phi(x,b)=\Phi(y,b') \end{align*} for elements $x,y \in A-C$ and $b,b' \in B$. Equality of ordered pairs gives \begin{align*} a_x-b &= a_y-b', \\ b-c_x &= b'-c_y. \end{align*} Adding these equalities is the key point: the intermediate terms involving $b$ and $b'$ cancel. Since $G$ is abelian, we may regroup the sums without changing their value: \begin{align*} (a_x-b)+(b-c_x) &=(a_y-b')+(b'-c_y), \\ a_x-c_x &=a_y-c_y. \end{align*} By the chosen representations, $a_x-c_x=x$ and $a_y-c_y=y$, so $x=y$. Once $x=y$, the chosen pair attached to $x$ is the same chosen pair attached to $y$, hence $a_x=a_y$ and $c_x=c_y$. Returning to the equality $a_x-b=a_y-b'$, we obtain $b=b'$ by cancellation in the group. Therefore $(x,b)=(y,b')$, and $\Phi$ is injective. [/guided] [/step] [step:Count the injection to obtain the cardinality inequality] Because $\Phi$ is an injective map from the finite set $(A-C)\times B$ into the finite set $(A-B)\times(B-C)$, cardinalities satisfy \begin{align*} |(A-C)\times B| \leq |(A-B)\times(B-C)|. \end{align*} Using the product formula for finite Cartesian products, \begin{align*} |A-C|\,|B| \leq |A-B|\,|B-C|. \end{align*} Since $B$ is non-empty, $|B|>0$, and division by $|B|$ gives \begin{align*} |A-C| \leq \frac{|A-B|\,|B-C|}{|B|}. \end{align*} [/step] [step:Convert the cardinality inequality into the distance inequality] For finite non-empty subsets $X,Y \subset G$, define \begin{align*} d(X,Y):=\log \frac{|X-Y|}{|X|^{1/2}|Y|^{1/2}}. \end{align*} The cardinality inequality gives \begin{align*} |A-C| \leq \frac{|A-B|\,|B-C|}{|B|}. \end{align*} Dividing both sides by $|A|^{1/2}|C|^{1/2}$ yields \begin{align*} \frac{|A-C|}{|A|^{1/2}|C|^{1/2}} \leq \frac{|A-B|\,|B-C|}{|A|^{1/2}|B|\,|C|^{1/2}}. \end{align*} Since \begin{align*} |A|^{1/2}|B|\,|C|^{1/2} = \left(|A|^{1/2}|B|^{1/2}\right) \left(|B|^{1/2}|C|^{1/2}\right), \end{align*} we have \begin{align*} \frac{|A-B|\,|B-C|}{|A|^{1/2}|B|\,|C|^{1/2}} = \frac{|A-B|}{|A|^{1/2}|B|^{1/2}} \cdot \frac{|B-C|}{|B|^{1/2}|C|^{1/2}}. \end{align*} All quantities are positive because $A,B,C$ are finite non-empty. Taking logarithms and using that $\log$ is increasing gives \begin{align*} d(A,C) \leq d(A,B)+d(B,C). \end{align*} This is the stated distance form. [/step]