[proofplan]
This entry records the finite-field sum-product theorem of Bourgain, Katz, and Tao, not an independent reconstruction of its incidence-geometric proof. The mathematical input is the external result of Bourgain--Katz--Tao, [A sum-product estimate in finite fields, and applications](https://doi.org/10.1007/s00222-003-0306-9), in the subfield-avoidance formulation. We spell out the avoidance hypothesis, choose the constants supplied by that external theorem, and verify that the present hypotheses match its hypotheses exactly.
[/proofplan]
[step:Define the additive and multiplicative expansion sets]
Let $F$ be a finite field, and let $A \subset F$ satisfy the hypotheses of the theorem. Define the sum set $A+A \subset F$ and the product set $AA \subset F$ by
\begin{align*}
A+A &= \{a_1+a_2 : a_1 \in A,\ a_2 \in A\}, \\
AA &= \{a_1a_2 : a_1 \in A,\ a_2 \in A\}.
\end{align*}
Because $C<|A|$, after increasing $C$ if necessary we may assume $|A|\geq 2$, so both sets are nonempty finite subsets of $F$.
[/step]
[step:Record the external Bourgain Katz Tao estimate and verify its hypotheses]
We use the external Bourgain--Katz--Tao finite-field sum-product theorem from Bourgain, Katz, and Tao, [A sum-product estimate in finite fields, and applications](https://doi.org/10.1007/s00222-003-0306-9). In the subfield-avoidance formulation needed here, it states that for every $\eta>0$ there exist constants $\delta_{\mathrm{BKT}}=\delta_{\mathrm{BKT}}(\eta)>0$ and $C_{\mathrm{BKT}}=C_{\mathrm{BKT}}(\eta)>0$ such that, whenever $F$ is a finite field and $A \subset F$ satisfies
\begin{align*}
C_{\mathrm{BKT}} < |A| < |F|^{1-\eta}
\end{align*}
and for every proper subfield $K \subsetneq F$ and every $x \in F$, $y \in F\setminus\{0\}$,
\begin{align*}
|A \cap (x+yK)| \leq \max\{|K|^{1/2}, |A|^{1-\eta}\},
\end{align*}
one has
\begin{align*}
\max(|A+A|,|AA|) \geq |A|^{1+\delta_{\mathrm{BKT}}}.
\end{align*}
This is a previously established external theorem, so the present argument is a specialization and normalization of constants rather than a proof of the incidence estimate itself. Choose
\begin{align*}
\delta &= \delta_{\mathrm{BKT}}, \\
C &= \max\{C_{\mathrm{BKT}},2\}.
\end{align*}
Then $C<|A|$ implies $C_{\mathrm{BKT}}<|A|$, while the upper bound $|A|<|F|^{1-\eta}$ and the displayed subfield-avoidance condition are hypotheses of the present theorem.
[guided]
The central point is provenance: we are not deriving the Bourgain--Katz--Tao incidence-combinatorial estimate inside this entry. We are invoking it as an external theorem, namely Bourgain, Katz, and Tao, [A sum-product estimate in finite fields, and applications](https://doi.org/10.1007/s00222-003-0306-9). In the formulation used here, that theorem says that for each fixed $\eta>0$ there are constants $\delta_{\mathrm{BKT}}(\eta)>0$ and $C_{\mathrm{BKT}}(\eta)>0$ such that every finite field $F$ and every subset $A \subset F$ satisfying
\begin{align*}
C_{\mathrm{BKT}} < |A| < |F|^{1-\eta}
\end{align*}
and, for every proper subfield $K \subsetneq F$ and every affine copy $x+yK$ with $x \in F$ and $y \in F\setminus\{0\}$,
\begin{align*}
|A \cap (x+yK)| \leq \max\{|K|^{1/2}, |A|^{1-\eta}\},
\end{align*}
must obey
\begin{align*}
\max(|A+A|,|AA|) \geq |A|^{1+\delta_{\mathrm{BKT}}}.
\end{align*}
We now check that the hypotheses of this external theorem match the hypotheses in the present statement. The upper size condition $|A|<|F|^{1-\eta}$ is assumed directly. The subfield-avoidance condition is also assumed directly: it is the displayed condition for every proper subfield $K \subsetneq F$ and every affine copy $x+yK$ with $y \neq 0$. The only normalization issue is the lower threshold, so define
\begin{align*}
\delta &= \delta_{\mathrm{BKT}}, \\
C &= \max\{C_{\mathrm{BKT}},2\}.
\end{align*}
Then $C<|A|$ implies $C_{\mathrm{BKT}}<|A|$, and the additional inequality $2<|A|$ only excludes degenerate one-point cases. Thus all assumptions of the cited Bourgain--Katz--Tao theorem are satisfied.
[/guided]
[/step]
[step:Read off the required lower bound]
The conclusion of the cited estimate gives
\begin{align*}
\max(|A+A|,|AA|) \geq |A|^{1+\delta_{\mathrm{BKT}}} = |A|^{1+\delta}.
\end{align*}
This is the claimed inequality, with $\delta>0$ and $C>0$ depending only on $\eta$.
[/step]
