[proofplan] We count additive quadruples by first choosing the ordered triple $(a_1,a_2,a_3) \in A^3$. Once this triple is fixed, the equation $a_1+a_2=a_3+a_4$ determines at most one possible value of $a_4$, namely $a_1+a_2-a_3$. Therefore the set of additive quadruples injects into $A^3$ by forgetting the fourth coordinate, giving the bound $E^+(A)\le |A|^3$. [/proofplan] [step:Define the set of additive quadruples to be counted] Let $Q \subset A^4$ denote the set of additive quadruples in $A$, defined by \begin{align*} Q := \left\{(a_1,a_2,a_3,a_4) \in A^4 : a_1+a_2=a_3+a_4\right\}. \end{align*} By the definition of additive energy, \begin{align*} E^+(A)=|Q|. \end{align*} [/step] [step:Show that the first three coordinates determine the fourth] Define the coordinate projection \begin{align*} \pi: Q &\to A^3 \\ (a_1,a_2,a_3,a_4) &\mapsto (a_1,a_2,a_3). \end{align*} We prove that $\pi$ is injective. Let $(a_1,a_2,a_3,a_4) \in Q$ and $(a_1,a_2,a_3,a_4') \in Q$ have the same first three coordinates. Since both quadruples lie in $Q$, we have \begin{align*} a_1+a_2 &= a_3+a_4, \\ a_1+a_2 &= a_3+a_4'. \end{align*} Cancelling $a_3$ from the two equalities in the abelian group $G$ gives \begin{align*} a_4=a_4'. \end{align*} Thus $\pi$ is injective. Equivalently, for a fixed ordered triple $(a_1,a_2,a_3) \in A^3$, any fourth coordinate satisfying the additive relation must be \begin{align*} a_4 = a_1+a_2-a_3, \end{align*} and this contributes a quadruple exactly when this element belongs to $A$. [/step] [step:Compare cardinalities to obtain the energy bound] Since $A$ is finite, the Cartesian product $A^3$ is finite and has cardinality \begin{align*} |A^3|=|A|^3. \end{align*} The injective map $\pi: Q \to A^3$ therefore gives \begin{align*} E^+(A)=|Q| \le |A^3|=|A|^3. \end{align*} This is the desired upper bound. [/step]