[proofplan]
We prove the corrected Elekes incidence estimate by the standard Szemerédi-Trotter argument. After discarding $0$ if necessary, we build $|B|^2$ distinct affine lines from non-zero elements of a large subset $B \subset A$ and count the incidences made by the points $(b+c,ac)$. The Szemerédi-Trotter Theorem converts the lower incidence count $|B|^3$ into the lower bound $|B+B|\,|BB| \ge C_0|B|^{5/2}$, and the elementary inclusions $B+B \subset A+A$ and $BB \subset AA$ transfer the estimate back to $A$ with an explicit absolute constant.
[/proofplan]
[step:Discard zero while retaining a large non-zero subset]
Let $N := |A|$ denote the cardinality of $A$, and define
\begin{align*}
B := A \setminus \{0\}.
\end{align*}
If $|B| \ge N/2$, we keep this set $B$. If $|B| < N/2$, then $N=1$ because removing a single element can reduce the size of $A$ below $N/2$ only when $N<2$. In this exceptional case $|A+A| \ge 1$ and $|AA| \ge 1$, so
\begin{align*}
|A+A| |AA| \ge 1 = |A|^{5/2}.
\end{align*}
Thus the corrected theorem holds in the exceptional case with any $c \le 1$. For the remainder of the proof assume $M := |B| \ge N/2$.
[/step]
[step:Construct the point set and the family of distinct lines]
Define the finite point set
\begin{align*}
P := (B+B) \times (BB) \subset \mathbb{R}^2,
\end{align*}
where $B+B := \{b+c : b,c \in B\}$ and $BB := \{ac : a,c \in B\}$. For each ordered pair $(a,b) \in B \times B$, define the affine line
\begin{align*}
\ell_{a,b} := \{(x,y) \in \mathbb{R}^2 : y = a(x-b)\}.
\end{align*}
Let
\begin{align*}
\mathcal{L} := \{\ell_{a,b} : (a,b) \in B \times B\}
\end{align*}
denote the resulting line family. Since every $a \in B$ is non-zero, the map $(a,b) \mapsto \ell_{a,b}$ is injective: if $\ell_{a,b}=\ell_{a',b'}$, equality of slopes gives $a=a'$, and then equality of intercepts gives $-ab=-ab'$, hence $b=b'$. Therefore
\begin{align*}
|\mathcal{L}| = M^2.
\end{align*}
[/step]
[step:Count at least $M^3$ incidences]
Let $I(P,\mathcal{L})$ be the number of incidences between $P$ and $\mathcal{L}$, namely
\begin{align*}
I(P,\mathcal{L}) := |\{(p,\ell) \in P \times \mathcal{L} : p \in \ell\}|.
\end{align*}
For every triple $(a,b,c) \in B \times B \times B$, define the point
\begin{align*}
p_{a,b,c} := (b+c,ac) \in \mathbb{R}^2.
\end{align*}
Because $b+c \in B+B$ and $ac \in BB$, we have $p_{a,b,c} \in P$. Moreover
\begin{align*}
ac = a((b+c)-b),
\end{align*}
so $p_{a,b,c} \in \ell_{a,b}$. For fixed $(a,b)$, the points $p_{a,b,c}$ are distinct as $c$ varies over $B$, since their first coordinates $b+c$ are distinct. Hence each of the $M^2$ lines contributes at least $M$ incidences, and therefore
\begin{align*}
I(P,\mathcal{L}) \ge M^3.
\end{align*}
[/step]
[step:Apply Szemerédi-Trotter to force the Elekes product bound]
By the Szemerédi-Trotter Theorem, there is an absolute constant $K>0$ such that for every finite point set $P \subset \mathbb{R}^2$ and every finite family of affine lines $\mathcal{L}$ in $\mathbb{R}^2$,
\begin{align*}
I(P,\mathcal{L}) \le K\left(|P|^{2/3}|\mathcal{L}|^{2/3}+|P|+|\mathcal{L}|\right).
\end{align*}
The sets $P$ and $\mathcal{L}$ constructed above are finite, so the theorem applies. Since $P=(B+B)\times(BB)$ and $|\mathcal{L}|=M^2$, the lower and upper incidence bounds give
\begin{align*}
M^3
&\le K\left(|P|^{2/3}M^{4/3}+|P|+M^2\right).
\end{align*}
Also $|P|=|B+B|\,|BB| \ge M^2$, because for any fixed $b_0 \in B$ the maps $b \mapsto b+b_0$ and $b \mapsto bb_0$ from $B$ into $B+B$ and $BB$, respectively, are injective; the second map is injective because $b_0 \ne 0$. Hence $M^2 \le |P|$, and therefore
\begin{align*}
M^3
&\le K\left(|P|^{2/3}M^{4/3}+2|P|\right).
\end{align*}
If $2K|P| \ge M^3/2$, then
\begin{align*}
|P| \ge \frac{M^3}{4K} \ge \frac{M^{5/2}}{4K},
\end{align*}
because $M \ge 1$. If instead $2K|P| < M^3/2$, subtracting this term from the preceding incidence inequality gives
\begin{align*}
\frac{M^3}{2} \le K |P|^{2/3}M^{4/3},
\end{align*}
and hence
\begin{align*}
|P| \ge (2K)^{-3/2}M^{5/2}.
\end{align*}
Define
\begin{align*}
C_0 := \min\{(4K)^{-1},(2K)^{-3/2}\}.
\end{align*}
Both cases give
\begin{align*}
|B+B|\,|BB| = |P| \ge C_0 M^{5/2}.
\end{align*}
[/step]
[step:Transfer the bound from $B$ back to $A$ and derive the maximum estimate]
Since $B \subset A$, we have $B+B \subset A+A$ and $BB \subset AA$. Therefore
\begin{align*}
|A+A| |AA|
&\ge |B+B| |BB| \\
&\ge C_0 M^{5/2} \\
&\ge C_0 2^{-5/2} N^{5/2}.
\end{align*}
Define the absolute constant
\begin{align*}
c := C_0 2^{-5/2}.
\end{align*}
This proves
\begin{align*}
|A+A| |AA| \ge c|A|^{5/2}.
\end{align*}
Finally, let $Q := \max(|A+A|,|AA|)$. Since $|A+A| \le Q$ and $|AA| \le Q$, we have
\begin{align*}
c|A|^{5/2} \le |A+A| |AA| \le Q^2.
\end{align*}
Taking square roots gives
\begin{align*}
\max(|A+A|,|AA|) = Q \ge c^{1/2}|A|^{5/4}.
\end{align*}
This is the stated consequence.
[/step]
