[proofplan] We encode the additive structure of $A$ and $B$ by the representation function $r_{A+B}$, which counts how many pairs $(a,b) \in A \times B$ produce each sum $x \in A+B$. The total mass of this function is $|A||B|$, while its second moment is exactly the additive energy $E^+(A,B)$. A finite Cauchy-Schwarz estimate, proved directly from a nonnegative square identity, then forces the second moment to be at least the square of the total mass divided by the size of the support. [/proofplan] [step:Define the representation function on the sumset] Define the representation function \begin{align*} r_{A+B}: A+B &\to \mathbb{N} \\ x &\mapsto \left|\{(a,b) \in A \times B : a+b=x\}\right|. \end{align*} Since $A$ and $B$ are finite and nonempty, the set $A+B$ is finite and nonempty. Every pair $(a,b) \in A \times B$ contributes to exactly one value of $r_{A+B}$, namely the value at $a+b \in A+B$. Therefore \begin{align*} \sum_{x \in A+B} r_{A+B}(x) = |A||B|. \end{align*} [/step] [step:Identify additive energy with the second moment of the representation function] For each $x \in A+B$, the number $r_{A+B}(x)^2$ counts ordered pairs of representations of $x$, namely ordered pairs \begin{align*} ((a_1,b_1),(a_2,b_2)) \in (A \times B) \times (A \times B) \end{align*} such that \begin{align*} a_1+b_1=x \qquad\text{and}\qquad a_2+b_2=x. \end{align*} Equivalently, it counts quadruples $(a_1,b_1,a_2,b_2) \in A \times B \times A \times B$ satisfying $a_1+b_1=a_2+b_2=x$. Summing over all $x \in A+B$ partitions the quadruples counted by $E^+(A,B)$ according to their common sum. Hence \begin{align*} E^+(A,B)=\sum_{x \in A+B} r_{A+B}(x)^2. \end{align*} [/step] [step:Apply the finite square identity to bound the second moment from below] Let $S:=A+B$ and let $f:S \to \mathbb{R}$ be the function $f(x):=r_{A+B}(x)$. Since $S$ is finite, the following finite identity is valid: \begin{align*} 0 &\leq \sum_{x \in S}\sum_{y \in S} (f(x)-f(y))^2 \\ &= \sum_{x \in S}\sum_{y \in S} \left(f(x)^2-2f(x)f(y)+f(y)^2\right) \\ &= 2|S|\sum_{x \in S} f(x)^2 - 2\left(\sum_{x \in S} f(x)\right)^2. \end{align*} Dividing by $2$ and rearranging gives \begin{align*} \left(\sum_{x \in S} f(x)\right)^2 \leq |S|\sum_{x \in S} f(x)^2. \end{align*} Substituting $S=A+B$ and $f=r_{A+B}$, we obtain \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*} [/step] [step:Substitute the mass and energy identities to obtain the lower bound] Using \begin{align*} \sum_{x \in A+B} r_{A+B}(x)=|A||B| \end{align*} and \begin{align*} E^+(A,B)=\sum_{x \in A+B} r_{A+B}(x)^2, \end{align*} the preceding inequality becomes \begin{align*} |A|^2|B|^2 \leq |A+B|\,E^+(A,B). \end{align*} Since $A+B$ is nonempty, $|A+B|>0$, so division by $|A+B|$ gives \begin{align*} E^+(A,B) \geq \frac{|A|^2 |B|^2}{|A+B|}. \end{align*} This is the claimed lower bound. [/step]