[proofplan] We encode the number of representations of an element $x \in G$ as a sum $a+b$ by a representation function $r_{A+B}$. Its total mass is $|A||B|$, while its second moment is exactly the additive energy $E^+(A,B)$. Applying the finite Cauchy--Schwarz inequality to this function on its support $A+B$ gives the desired lower bound. The special case follows by setting $B=A$. [/proofplan] [step:Define the representation function and compute its total mass] Define the representation function \begin{align*} r_{A+B}: G &\to \mathbb{N} \cup \{0\} \\ x &\mapsto \left|\{(a,b) \in A \times B : a+b=x\}\right|. \end{align*} Its support is exactly $A+B$, because $r_{A+B}(x)>0$ iff $x=a+b$ for some $(a,b) \in A \times B$. Since the fibers of the map \begin{align*} \sigma: A \times B &\to G \\ (a,b) &\mapsto a+b \end{align*} partition $A \times B$, we have \begin{align*} \sum_{x \in A+B} r_{A+B}(x) = |A \times B| = |A||B|. \end{align*} [guided] The function $r_{A+B}$ records multiplicity: it does not merely say whether $x$ lies in $A+B$, but counts how many ordered pairs $(a,b) \in A \times B$ produce $x$. Formally, define \begin{align*} r_{A+B}: G &\to \mathbb{N} \cup \{0\} \\ x &\mapsto \left|\{(a,b) \in A \times B : a+b=x\}\right|. \end{align*} The support of this function is $A+B$, because $r_{A+B}(x)>0$ exactly when $x$ has at least one representation $x=a+b$ with $a \in A$ and $b \in B$. Now define the addition map \begin{align*} \sigma: A \times B &\to G \\ (a,b) &\mapsto a+b. \end{align*} For each $x \in A+B$, the fiber $\sigma^{-1}(\{x\})$ is precisely the set counted by $r_{A+B}(x)$. These fibers are pairwise disjoint and their union is $A \times B$. Therefore summing their cardinalities gives \begin{align*} \sum_{x \in A+B} r_{A+B}(x) = \sum_{x \in A+B} |\sigma^{-1}(\{x\})| = |A \times B| = |A||B|. \end{align*} This identity is the input to Cauchy--Schwarz: the total number of representations is fixed, and we will compare it to the second moment of the representation counts. [/guided] [/step] [step:Identify the second moment with additive energy] For each $x \in A+B$, define \begin{align*} R_x := \{(a,b) \in A \times B : a+b=x\}. \end{align*} Then $r_{A+B}(x)=|R_x|$, and \begin{align*} \sum_{x \in A+B} r_{A+B}(x)^2 &= \sum_{x \in A+B} |R_x|^2 \\ &= \left|\{((a_1,b_1),(a_2,b_2)) \in (A \times B)^2 : a_1+b_1=a_2+b_2\}\right| \\ &= E^+(A,B). \end{align*} [guided] The square $r_{A+B}(x)^2$ has a direct counting meaning. For a fixed $x \in A+B$, define \begin{align*} R_x := \{(a,b) \in A \times B : a+b=x\}. \end{align*} Then $r_{A+B}(x)=|R_x|$, so $r_{A+B}(x)^2$ counts ordered pairs of representations of the same element $x$: \begin{align*} R_x \times R_x = \{((a_1,b_1),(a_2,b_2)) \in (A \times B)^2 : a_1+b_1=x,\ a_2+b_2=x\}. \end{align*} Summing over $x \in A+B$ counts every ordered pair of representations with equal sums exactly once, namely at the common value $x=a_1+b_1=a_2+b_2$. Hence \begin{align*} \sum_{x \in A+B} r_{A+B}(x)^2 &= \sum_{x \in A+B} |R_x|^2 \\ &= \left|\{((a_1,b_1),(a_2,b_2)) \in (A \times B)^2 : a_1+b_1=a_2+b_2\}\right| \\ &= E^+(A,B). \end{align*} Thus the additive energy is the second moment of the representation function. [/guided] [/step] [step:Apply finite Cauchy--Schwarz on the support $A+B$] We use the finite inequality \begin{align*} \left(\sum_{x \in S} u(x)\right)^2 \leq |S| \sum_{x \in S} u(x)^2 \end{align*} for every finite nonempty set $S$ and every function $u:S \to \mathbb{R}$. Indeed, \begin{align*} 0 &\leq \sum_{x \in S}\sum_{y \in S} (u(x)-u(y))^2 \\ &= 2|S|\sum_{x \in S}u(x)^2 - 2\left(\sum_{x \in S}u(x)\right)^2. \end{align*} Apply this inequality with $S=A+B$ and $u=r_{A+B}|_{A+B}$. Since $A$ and $B$ are nonempty, $A+B$ is finite and nonempty. Therefore \begin{align*} |A|^2|B|^2 &= \left(\sum_{x \in A+B} r_{A+B}(x)\right)^2 \\ &\leq |A+B| \sum_{x \in A+B} r_{A+B}(x)^2 \\ &= |A+B| E^+(A,B). \end{align*} Dividing by the positive integer $|A+B|$ gives \begin{align*} E^+(A,B) \geq \frac{|A|^2|B|^2}{|A+B|}. \end{align*} [guided] We need an inequality that converts the fixed total mass of $r_{A+B}$ into a lower bound for its second moment. The required finite Cauchy--Schwarz inequality says that for every finite nonempty set $S$ and every function $u:S \to \mathbb{R}$, \begin{align*} \left(\sum_{x \in S} u(x)\right)^2 \leq |S| \sum_{x \in S} u(x)^2. \end{align*} For completeness, here is a proof in the finite setting. Since every square is nonnegative, \begin{align*} 0 &\leq \sum_{x \in S}\sum_{y \in S} (u(x)-u(y))^2 \\ &= \sum_{x \in S}\sum_{y \in S} \left(u(x)^2 - 2u(x)u(y) + u(y)^2\right) \\ &= |S|\sum_{x \in S}u(x)^2 - 2\left(\sum_{x \in S}u(x)\right)\left(\sum_{y \in S}u(y)\right) + |S|\sum_{y \in S}u(y)^2 \\ &= 2|S|\sum_{x \in S}u(x)^2 - 2\left(\sum_{x \in S}u(x)\right)^2. \end{align*} Rearranging gives the stated inequality. Now take $S=A+B$ and let $u:S \to \mathbb{R}$ be the restriction $u=r_{A+B}|_{A+B}$. The set $A+B$ is finite because $A$ and $B$ are finite, and it is nonempty because $A$ and $B$ are nonempty. Hence the finite inequality applies: \begin{align*} \left(\sum_{x \in A+B} r_{A+B}(x)\right)^2 \leq |A+B| \sum_{x \in A+B} r_{A+B}(x)^2. \end{align*} Using the total-mass identity from the first step and the second-moment identity from the second step, this becomes \begin{align*} |A|^2|B|^2 \leq |A+B| E^+(A,B). \end{align*} Since $|A+B|>0$, division gives \begin{align*} E^+(A,B) \geq \frac{|A|^2|B|^2}{|A+B|}. \end{align*} [/guided] [/step] [step:Specialize to $B=A$] Set $B=A$. Then $A+B=A+A$ and $E^+(A,B)=E^+(A,A)=E^+(A)$ by definition. The preceding inequality gives \begin{align*} E^+(A) \geq \frac{|A|^4}{|A+A|}. \end{align*} This proves both assertions. [/step]