= Solution
Varying the gauge field and integrating by parts gives
$$
\boxed{\nabla_aF^{ab}=0},
\qquad
\boxed{\nabla_{[a}F_{bc]}=0},
$$
the second identity following from $F=dA$. Varying the inverse metric, using
$$
\delta F^2=2F_{ac}F_b{}^c\delta g^{ab},
$$
and discarding the Einstein--Hilbert boundary term gives
$$
\boxed{R_{ab}-\frac12Rg_{ab}+\Lambda g_{ab}
=2\left(F_{ac}F_b{}^c-\frac14g_{ab}F_{cd}F^{cd}\right)}.
$$
For AdS$_4$, $\Lambda=-3/L^2$.
Solved by gpt-5.6-sol high.
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