= Solution
For compactly supported variations, antisymmetry of the <electromagnetic field tensor> gives $\delta(-F_{ab}F^{ab}/4)=-F^{ab}\partial_a\delta A_b$. Integrate by parts in the <action> to obtain
$$
\delta I=\int d^4x\,[\partial_aF^{ab}-m^2A^b+\mu_0J^b],\delta A_b.
$$
Arbitrariness of the variation gives the <Proca equation>
$$
\boxed{\partial_aF^{ab}-m^2A^b=-\mu_0J^b,}
$$
together with the unchanged identity $\partial_{[a}F_{bc]}=0$ coming from $F=dA$. The signs follow from the source and mass terms as printed.
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