[proofplan] This proof records the standard corollary of the independently established Bourgain-Katz-Tao incidence estimate over prime fields: for each fixed density gap $\varepsilon>0$, that estimate gives a power improvement for $|A+A|+|A\cdot A|$ whenever $p$ is sufficiently large and $1<|A|<p^{1-\varepsilon}$. We then convert the lower bound for the sum of the two expansion quantities into a lower bound for their maximum, losing only by replacing the exponent with a smaller positive exponent. The finitely many cardinalities not covered after this exponent loss are handled uniformly by Cauchy-Davenport, so the final exponent depends only on $\varepsilon$. [/proofplan] [step:Apply the independently established Bourgain-Katz-Tao incidence estimate] Let $N := |A|$ denote the cardinality of $A$. Define the sumset $S \subset \mathbb F_p$ and product set $P \subset \mathbb F_p$ by \begin{align*} S &:= A+A = \{a+b : a,b \in A\}, \\ P &:= A\cdot A = \{ab : a,b \in A\}. \end{align*} We use the following independently proved Bourgain-Katz-Tao incidence estimate over prime fields: for each fixed $\varepsilon>0$ there exist constants $p_0(\varepsilon) \in \mathbb N$ and $\kappa(\varepsilon)>0$ such that, for every prime $p \ge p_0(\varepsilon)$ and every subset $A \subset \mathbb F_p$ satisfying $1<|A|<p^{1-\varepsilon}$, \begin{align*} |A+A|+|A\cdot A| \ge |A|^{1+\kappa(\varepsilon)}. \end{align*} The hypotheses of this incidence estimate are satisfied: $p$ is prime, $p \ge p_0(\varepsilon)$, $A \subset \mathbb F_p$, $1<N<p^{1-\varepsilon}$, and $\varepsilon>0$ is fixed. Therefore \begin{align*} |S|+|P| \ge N^{1+\kappa(\varepsilon)}. \end{align*} [/step] [step:Convert the sum bound into a maximum bound for large sets] Define \begin{align*} M := \max\{|S|,|P|\}. \end{align*} Since $|S| \le M$ and $|P| \le M$, we have $|S|+|P| \le 2M$. Combining this with the preceding estimate gives \begin{align*} M \ge \frac{1}{2}N^{1+\kappa(\varepsilon)}. \end{align*} For every integer $N$ satisfying $N^{\kappa(\varepsilon)/2} \ge 2$, this implies \begin{align*} M \ge N^{1+\kappa(\varepsilon)/2}. \end{align*} [/step] [step:Handle the finitely many small cardinalities] Let $K := \kappa(\varepsilon)$ and define the finite set of possible small cardinalities \begin{align*} \mathcal N_K := \{N \in \mathbb N : N \ge 2 \text{ and } N^{K/2}<2\}. \end{align*} For $N \in \mathcal N_K$, the Cauchy-Davenport theorem gives \begin{align*} |A+A| \ge \min\{p,2N-1\}. \end{align*} Because $A \subset \mathbb F_p$ has $N<p$, we have $p \ge N+1$, and also $2N-1 \ge N+1$ for $N\ge2$. Hence \begin{align*} |A+A| \ge N+1. \end{align*} Define \begin{align*} c_0(K) := \min_{N \in \mathcal N_K}\left(\frac{\log(N+1)}{\log N}-1\right) \end{align*} if $\mathcal N_K$ is nonempty, and define $c_0(K):=K/2$ if $\mathcal N_K$ is empty. This constant is positive because $\mathcal N_K$ is finite and $N+1>N$ for each $N \in \mathcal N_K$. Therefore, for every $N \in \mathcal N_K$, \begin{align*} M \ge |A+A| \ge N+1 \ge N^{1+c_0(K)}. \end{align*} [/step] [step:Choose a uniform exponent and conclude the theorem] Define \begin{align*} c(\varepsilon) := \min\left\{\frac{\kappa(\varepsilon)}{2}, c_0(\kappa(\varepsilon))\right\}. \end{align*} This number is positive and depends only on $\varepsilon$. If $N^{\kappa(\varepsilon)/2} \ge 2$, the large-cardinality estimate gives \begin{align*} M \ge N^{1+\kappa(\varepsilon)/2} \ge N^{1+c(\varepsilon)}. \end{align*} If $N^{\kappa(\varepsilon)/2}<2$, then $N \in \mathcal N_{\kappa(\varepsilon)}$, and the small-cardinality estimate gives \begin{align*} M \ge N^{1+c_0(\kappa(\varepsilon))} \ge N^{1+c(\varepsilon)}. \end{align*} Since $M=\max\{|A+A|,|A\cdot A|\}$ and $N=|A|$, this proves \begin{align*} \max\{|A+A|,|A\cdot A|\} \ge |A|^{1+c(\varepsilon)}. \end{align*} [/step]