= Solution
Use $u^\lambda u_\lambda=-c^2$ and a constant <Minkowski metric>. Vary the velocity-dependent square root \b[before] imposing the proper-time normalization. Equivalently, vary a reparametrization-invariant <worldline> action first and then choose <proper time>. Its canonical covector momentum is
$$
\frac{\partial\mathcal L}{\partial u^a}
=mc\frac{u_a}{\sqrt{-u^2}}+qA_a
\ \longrightarrow\ m u_a+qA_a.
$$
The coordinate derivative at fixed velocity is $\partial\mathcal L/\partial x^a=q u^\lambda\partial_aA_\lambda$. The <Euler-Lagrange equation> and the <chain rule> from <2/a> therefore give
$$
m\dot u_a+q u^\lambda\partial_\lambda A_a
=q u^\lambda\partial_a A_\lambda.
$$
Define the antisymmetric field tensor by
$$
\boxed{F_{ab}=\partial_aA_b-\partial_bA_a,\qquad
F^{ab}=\partial^aA^b-\partial^bA^a.}
$$
The particle equation becomes
$$
\boxed{m\frac{du_a}{d\tau}=qF_{a\lambda}u^\lambda.}
$$
Antisymmetry ensures $u^aF_{a\lambda}u^\lambda=0$, so the force is orthogonal to the <four-velocity> and preserves $u^2=-c^2$. The <worldline variation before proper-time normalization> is essential: substituting $\sqrt{-u^2}=c$ into the <Lagrangian> \b[before taking its velocity derivative] would turn the kinetic term into a constant and incorrectly remove the inertial momentum. The normalization is a property of the proper-time solution, not permission to discard the variational dependence.
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