[proofplan]
We group the possible nonzero values of the representation function $r_{A+B}$ into dyadic ranges. Since no element of $G$ can have more than $\min\{|A|,|B|\}$ representations as a sum $a+b$, only logarithmically many dyadic ranges occur. The additive energy is the sum of $r_{A+B}(x)^2$ over these ranges, so one dyadic range contributes at least the average amount. On that range, $r_{A+B}(x)$ is comparable to the lower endpoint $\tau$, giving the desired lower bound for $|P|\tau^2$.
[/proofplan]
[step:Bound the possible representation counts]
Let
\begin{align*}
M:=\min\{|A|,|B|\}.
\end{align*}
Since $A$ and $B$ are nonempty, $M \geq 1$. For every $x \in G$, each representation $x=a+b$ is determined by its $A$-coordinate $a \in A$, and also by its $B$-coordinate $b \in B$. Therefore
\begin{align*}
0 \leq r_{A+B}(x) \leq M
\end{align*}
for every $x \in G$.
Moreover, $r_{A+B}(x)=0$ whenever $x \notin A+B$, so the energy sum is finite and may be written as
\begin{align*}
E=\sum_{x \in A+B} r_{A+B}(x)^2.
\end{align*}
[/step]
[step:Partition the nonzero representation counts into dyadic levels]
Let
\begin{align*}
L:=\lfloor \log_2 M \rfloor.
\end{align*}
For each integer $j$ with $0 \leq j \leq L$, define the dyadic level
\begin{align*}
P_j:=\{x \in A+B : 2^j \leq r_{A+B}(x) < 2^{j+1}\}.
\end{align*}
Every $x \in A+B$ has $r_{A+B}(x) \geq 1$, and by the previous step $r_{A+B}(x) \leq M$. Hence the sets $P_0,\dots,P_L$ form a disjoint partition of $A+B$.
Thus
\begin{align*}
E
&=\sum_{x \in A+B} r_{A+B}(x)^2 \\
&=\sum_{j=0}^{L}\sum_{x \in P_j} r_{A+B}(x)^2.
\end{align*}
[guided]
The purpose of the dyadic partition is to replace the variable quantity $r_{A+B}(x)$ by a single scale. We use the levels
\begin{align*}
P_j:=\{x \in A+B : 2^j \leq r_{A+B}(x) < 2^{j+1}\}
\end{align*}
for integers $j$ satisfying $0 \leq j \leq L$, where
\begin{align*}
L:=\lfloor \log_2 M \rfloor,
\qquad
M:=\min\{|A|,|B|\}.
\end{align*}
This range of indices covers all possible nonzero representation counts. Indeed, if $x \in A+B$, then $r_{A+B}(x) \geq 1=2^0$. Also, from the previous step, $r_{A+B}(x) \leq M$, so $r_{A+B}(x)$ lies in one of the intervals
\begin{align*}
[2^0,2^1),[2^1,2^2),\dots,[2^L,2^{L+1}).
\end{align*}
The intervals are disjoint, so the sets $P_0,\dots,P_L$ form a disjoint partition of $A+B$.
Since $r_{A+B}(x)=0$ outside $A+B$, the additive energy is
\begin{align*}
E
&=\sum_{x \in A+B} r_{A+B}(x)^2 \\
&=\sum_{j=0}^{L}\sum_{x \in P_j} r_{A+B}(x)^2.
\end{align*}
This rewrites the energy as a sum of contributions from logarithmically many popularity levels.
[/guided]
[/step]
[step:Choose a dyadic level with average energy contribution]
There are $L+1$ dyadic levels. Since the nonnegative numbers
\begin{align*}
E_j:=\sum_{x \in P_j} r_{A+B}(x)^2,
\qquad 0 \leq j \leq L,
\end{align*}
satisfy
\begin{align*}
\sum_{j=0}^{L} E_j=E,
\end{align*}
there exists an index $j_0 \in \{0,\dots,L\}$ such that
\begin{align*}
E_{j_0}\geq \frac{E}{L+1}.
\end{align*}
Define
\begin{align*}
\tau:=2^{j_0},
\qquad
P:=P_{j_0}.
\end{align*}
Then $\tau \geq 1$ and
\begin{align*}
P=\{x \in A+B : \tau \leq r_{A+B}(x) < 2\tau\}.
\end{align*}
[/step]
[step:Convert the energy contribution into the popular-sums bound]
For every $x \in P$, the definition of $P$ gives
\begin{align*}
r_{A+B}(x)<2\tau.
\end{align*}
Therefore
\begin{align*}
E_{j_0}
=\sum_{x \in P} r_{A+B}(x)^2
< \sum_{x \in P} (2\tau)^2
=4|P|\tau^2.
\end{align*}
Combining this with $E_{j_0}\geq E/(L+1)$ gives
\begin{align*}
|P|\tau^2
> \frac{E}{4(L+1)}.
\end{align*}
Finally,
\begin{align*}
L+1=\lfloor \log_2 M \rfloor+1 \leq \log_2(2M)
=\frac{\log(2M)}{\log 2}.
\end{align*}
Hence
\begin{align*}
|P|\tau^2
\geq \frac{E}{C\log(2M)}
\end{align*}
with the absolute constant
\begin{align*}
C:=\frac{4}{\log 2}.
\end{align*}
Since $M=\min\{|A|,|B|\}$, this is exactly
\begin{align*}
|P|\tau^2
\geq \frac{E}{C\log(2\min\{|A|,|B|\})}.
\end{align*}
This proves the theorem.
[/step]
