[proofplan]
We prove the stronger quantitative bound $\max\{|A+A|,|AA|\} \geq c|A|^{5/4}$ for an absolute constant $c>0$. The argument is Elekes's incidence-geometric proof: from every triple $(r,b,a)$ with $r \in A \setminus \{0\}$ and $a,b \in A$, we build an incidence between a point of $(A+A)\times AA$ and a line indexed by $(r,b)$. The Szemerédi-Trotter incidence theorem then bounds the number of such incidences from above, while the construction gives many incidences from below. Rearranging the resulting inequality gives the desired power saving.
[/proofplan]
[step:Handle the one-point case and define the incidence parameters]
Let $n := |A|$. If $n=1$, then $|A+A|=1$ and $|AA|=1$, so the theorem holds for every constant $c \leq 1$ and every $\delta>0$.
Assume henceforth that $n \geq 2$. Define
\begin{align*}
B := A \setminus \{0\}.
\end{align*}
Since at most one element of $A$ is zero, we have
\begin{align*}
|B| \geq n-1 \geq \frac{n}{2}.
\end{align*}
Let
\begin{align*}
S := |A+A|, \qquad P := |AA|, \qquad M := \max\{S,P\}.
\end{align*}
Then $S \leq M$, $P \leq M$, and hence $SP \leq M^2$.
Also $M \geq n$: choosing any fixed $a_0 \in A$, the map
\begin{align*}
\tau_{a_0}: A &\to A+A \\
a &\mapsto a+a_0
\end{align*}
is injective, so $|A+A| \geq |A|=n$.
[/step]
[step:Build points and lines encoding sums and products]
Define the finite point set
\begin{align*}
\mathcal{P} := (A+A) \times AA \subset \mathbb{R}^2.
\end{align*}
Thus
\begin{align*}
|\mathcal{P}| = |A+A|\,|AA| = SP \leq M^2.
\end{align*}
For each ordered pair $(r,b) \in B \times A$, define the affine line
\begin{align*}
\ell_{r,b} := \{(x,y) \in \mathbb{R}^2 : y = r(x-b)\}.
\end{align*}
Let
\begin{align*}
\mathcal{L} := \{\ell_{r,b} : (r,b) \in B \times A\}.
\end{align*}
The map $(r,b) \mapsto \ell_{r,b}$ is injective. Indeed, if $\ell_{r,b}=\ell_{r',b'}$, then the slopes are equal, so $r=r'$. Since $r \in B$, we have $r \neq 0$. Equality of the intercepts gives $-rb=-rb'$, hence $b=b'$. Therefore
\begin{align*}
|\mathcal{L}| = |B|\,|A| \leq n^2.
\end{align*}
For every triple $(r,b,a) \in B \times A \times A$, define the point
\begin{align*}
q_{r,b,a} := (a+b,ra) \in \mathbb{R}^2.
\end{align*}
Since $a+b \in A+A$ and $ra \in AA$, we have $q_{r,b,a} \in \mathcal{P}$. Moreover
\begin{align*}
ra = r((a+b)-b),
\end{align*}
so $q_{r,b,a} \in \ell_{r,b}$. Hence each triple $(r,b,a)$ gives an incidence between a point of $\mathcal{P}$ and a line of $\mathcal{L}$.
For fixed $(r,b)$, the map
\begin{align*}
\iota_{r,b}: A &\to \mathcal{P} \\
a &\mapsto (a+b,ra)
\end{align*}
is injective because $a+b=a'+b$ implies $a=a'$. Therefore the construction gives at least $n$ distinct incidences on each line $\ell_{r,b}$. Consequently the total incidence count
\begin{align*}
I(\mathcal{P},\mathcal{L}) := |\{(q,\ell) \in \mathcal{P}\times \mathcal{L} : q \in \ell\}|
\end{align*}
satisfies
\begin{align*}
I(\mathcal{P},\mathcal{L}) \geq |B|\,|A|\,|A| \geq \frac{n^3}{2}.
\end{align*}
[guided]
The point set is chosen so that its first coordinate records a sum and its second coordinate records a product:
\begin{align*}
\mathcal{P} := (A+A)\times AA.
\end{align*}
Thus a point of $\mathcal{P}$ has the form $(s,p)$ with $s=a_1+a_2$ and $p=a_3a_4$ for elements of $A$.
Now fix $r \in B=A\setminus\{0\}$ and $b \in A$. Define
\begin{align*}
\ell_{r,b} := \{(x,y) \in \mathbb{R}^2 : y=r(x-b)\}.
\end{align*}
The exclusion of $0$ from $B$ is used to make the line parameterization injective. If $\ell_{r,b}=\ell_{r',b'}$, equality of slopes gives $r=r'$. Equality of intercepts gives $-rb=-rb'$. Since $r\neq 0$, this implies $b=b'$. Hence the number of distinct lines is exactly
\begin{align*}
|\mathcal{L}|=|B|\,|A|.
\end{align*}
The incidence identity is the elementary equation
\begin{align*}
ra = r((a+b)-b).
\end{align*}
For each $a \in A$, the point
\begin{align*}
q_{r,b,a}:=(a+b,ra)
\end{align*}
belongs to $\mathcal{P}$ because $a+b \in A+A$ and $ra \in AA$. The displayed identity says exactly that $q_{r,b,a}$ lies on $\ell_{r,b}$.
For fixed $(r,b)$, distinct values of $a$ give distinct points, since $a+b=a'+b$ implies $a=a'$. Thus each line $\ell_{r,b}$ contributes at least $n$ incidences. Since there are $|B|\,|A|$ such lines and $|B|\geq n/2$, we obtain
\begin{align*}
I(\mathcal{P},\mathcal{L}) \geq |B|\,|A|\,|A| \geq \frac{n^3}{2}.
\end{align*}
[/guided]
[/step]
[step:Apply the Szemerédi-Trotter incidence theorem]
We use the Szemerédi-Trotter incidence theorem for points and lines in $\mathbb{R}^2$ (citing a result not yet in the wiki: Szemerédi-Trotter incidence theorem). It gives an absolute constant $K>0$ such that, for every finite point set $\mathcal{Q}\subset \mathbb{R}^2$ and every finite set of lines $\mathcal{M}$ in $\mathbb{R}^2$,
\begin{align*}
I(\mathcal{Q},\mathcal{M})
\leq
K\left(|\mathcal{Q}|^{2/3}|\mathcal{M}|^{2/3}+|\mathcal{Q}|+|\mathcal{M}|\right).
\end{align*}
The theorem applies because $\mathcal{P}$ and $\mathcal{L}$ are finite subsets of the Euclidean plane and of the set of affine lines in that plane.
Using $|\mathcal{P}| \leq M^2$ and $|\mathcal{L}| \leq n^2$, we obtain
\begin{align*}
\frac{n^3}{2}
\leq I(\mathcal{P},\mathcal{L})
\leq
K\left((M^2)^{2/3}(n^2)^{2/3}+M^2+n^2\right).
\end{align*}
Since $M \geq n$, we have $n^2 \leq M^2$, and therefore
\begin{align*}
\frac{n^3}{2}
\leq
K\left(M^{4/3}n^{4/3}+2M^2\right).
\end{align*}
[guided]
The incidence construction gives the lower bound
\begin{align*}
I(\mathcal{P},\mathcal{L}) \geq \frac{n^3}{2}.
\end{align*}
To contradict small sumset and product set sizes, we now need an upper bound on the same incidence count.
The Szemerédi-Trotter theorem applies to finite point sets and finite sets of affine lines in $\mathbb{R}^2$. Our point set
\begin{align*}
\mathcal{P}=(A+A)\times AA
\end{align*}
is finite because $A$ is finite, and our line set
\begin{align*}
\mathcal{L}=\{\ell_{r,b}:r\in B,\ b\in A\}
\end{align*}
is finite for the same reason. Hence there is an absolute constant $K>0$ such that
\begin{align*}
I(\mathcal{P},\mathcal{L})
\leq
K\left(|\mathcal{P}|^{2/3}|\mathcal{L}|^{2/3}+|\mathcal{P}|+|\mathcal{L}|\right).
\end{align*}
We have already bounded the two sizes:
\begin{align*}
|\mathcal{P}| = |A+A|\,|AA| = SP \leq M^2
\end{align*}
and
\begin{align*}
|\mathcal{L}|=|B|\,|A|\leq n^2.
\end{align*}
Substituting these estimates into Szemerédi-Trotter gives
\begin{align*}
I(\mathcal{P},\mathcal{L})
\leq
K\left((M^2)^{2/3}(n^2)^{2/3}+M^2+n^2\right)
=
K\left(M^{4/3}n^{4/3}+M^2+n^2\right).
\end{align*}
Finally, $M\geq n$ because $|A+A|\geq n$, so $n^2\leq M^2$. Combining the upper and lower incidence bounds gives
\begin{align*}
\frac{n^3}{2}
\leq
K\left(M^{4/3}n^{4/3}+2M^2\right).
\end{align*}
[/guided]
[/step]
[step:Rearrange the incidence inequality into a sum-product lower bound]
From
\begin{align*}
\frac{n^3}{2}
\leq
K\left(M^{4/3}n^{4/3}+2M^2\right),
\end{align*}
there are two cases.
If
\begin{align*}
2KM^2 \geq \frac{n^3}{4},
\end{align*}
then
\begin{align*}
M \geq \frac{1}{\sqrt{8K}}\,n^{3/2} \geq \frac{1}{\sqrt{8K}}\,n^{5/4},
\end{align*}
because $n \geq 1$.
Otherwise
\begin{align*}
2KM^2 < \frac{n^3}{4}.
\end{align*}
Subtracting this strict upper bound for the quadratic term from the incidence inequality yields
\begin{align*}
K M^{4/3}n^{4/3} \geq \frac{n^3}{4}.
\end{align*}
Dividing by $K n^{4/3}$ gives
\begin{align*}
M^{4/3} \geq \frac{1}{4K}n^{5/3}.
\end{align*}
Raising both sides to the power $3/4$ gives
\begin{align*}
M \geq \left(\frac{1}{4K}\right)^{3/4}n^{5/4}.
\end{align*}
Therefore, with
\begin{align*}
c := \min\left\{1,\frac{1}{\sqrt{8K}},\left(\frac{1}{4K}\right)^{3/4}\right\}>0,
\end{align*}
we have
\begin{align*}
M=\max\{|A+A|,|AA|\} \geq c n^{5/4}.
\end{align*}
Since $n=|A|$, this proves the theorem with $\delta=\frac{1}{4}$.
[guided]
The inequality obtained from incidences is
\begin{align*}
\frac{n^3}{2}
\leq
K\left(M^{4/3}n^{4/3}+2M^2\right).
\end{align*}
The right-hand side has two terms. At least one of them must be large enough to account for the lower bound $n^3/2$.
First suppose the quadratic term is already large:
\begin{align*}
2KM^2 \geq \frac{n^3}{4}.
\end{align*}
Then
\begin{align*}
M^2 \geq \frac{n^3}{8K},
\end{align*}
so
\begin{align*}
M \geq \frac{1}{\sqrt{8K}}n^{3/2}.
\end{align*}
Since $n\geq 1$, this implies the weaker but sufficient estimate
\begin{align*}
M \geq \frac{1}{\sqrt{8K}}n^{5/4}.
\end{align*}
Now suppose the quadratic term is not large:
\begin{align*}
2KM^2 < \frac{n^3}{4}.
\end{align*}
Then the remaining term must carry the incidence count. Substituting this bound into
\begin{align*}
\frac{n^3}{2}
\leq
K M^{4/3}n^{4/3}+2KM^2
\end{align*}
gives
\begin{align*}
\frac{n^3}{2}
<
K M^{4/3}n^{4/3}+\frac{n^3}{4}.
\end{align*}
Hence
\begin{align*}
K M^{4/3}n^{4/3} \geq \frac{n^3}{4}.
\end{align*}
Dividing by $K n^{4/3}$ gives
\begin{align*}
M^{4/3} \geq \frac{1}{4K}n^{5/3}.
\end{align*}
Taking the power $3/4$ on both sides yields
\begin{align*}
M \geq \left(\frac{1}{4K}\right)^{3/4}n^{5/4}.
\end{align*}
Both cases therefore give $M \geq c n^{5/4}$ after choosing
\begin{align*}
c := \min\left\{1,\frac{1}{\sqrt{8K}},\left(\frac{1}{4K}\right)^{3/4}\right\}.
\end{align*}
The extra entry $1$ in the minimum also covers the one-point case considered at the beginning. Since $M=\max\{|A+A|,|AA|\}$ and $n=|A|$, we obtain
\begin{align*}
\max\{|A+A|,|AA|\} \geq c |A|^{5/4}.
\end{align*}
Thus the stated Erdős-Szemerédi form holds with $\delta=1/4$.
[/guided]
[/step]
