Nonsymmetric Lax--Milgram is not a scalar minimisation principle

Existence and uniqueness in The Lax--Milgram theorem do not require symmetry: only boundedness and coercivity are used, and the contraction proof never symmetrises the form. Symmetry is, however, exactly what the energy characterisation of Symmetric Lax--Milgram is energy minimisation consumes. If $a$ is bounded and coercive but not symmetric, and $F$ is bounded and conjugate-linear, then the solution of $a(u,v)=F(v)$ is in general not a critical point, and not a minimiser, of $v\mapsto \frac{1}{2}\mathrm{Re}a(v,v)-\mathrm{Re}F(v)$: the Euler--Lagrange equation of that functional involves the symmetrised form $a_s=\frac{1}{2}(a+a^{\ast })$, whose operator is $\frac{1}{2}(A+A^{\ast })$, whereas the weak equation involves $A$. The two coincide exactly when $a$ is symmetric. The companion page's positive-definite-plus-skew counterexample and drift example exhibit the failure concretely. This is a remark: it records the boundary of the variational statement and is not used as a proof step.