[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*}[/step]

[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: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*}[/step]

[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: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 $\delta=\frac{1}{4}$.[/step]

[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]