= Solution
The real <Lax-Milgram theorem> applies to a <Hilbert space> $H$ and a <bounded bilinear form> $a$ satisfying
$$
|a(u,v)|\leq M\|u\|\|v\|,\qquad a(v,v)\geq\alpha\|v\|^2\quad(\alpha>0).
$$
For every bounded <linear functional> $L$ there is a unique $u\in H$ with
$$
\boxed{a(u,v)=L(v)\quad(v\in H),\qquad\|u\|\leq\alpha^{-1}\|L\|.}
$$
Symmetry of the <bilinear form> is not required.
By the <Riesz representation theorem>, write $a(u,v)=\langle Au,v\rangle$ and $L(v)=\langle g,v\rangle$. The operator $A$ is linear and bounded, with $\|A\|\leq M$. The <coercive bilinear form> bound and the <Cauchy-Schwarz inequality> imply
$$
\alpha\|u\|^2\leq\langle Au,u\rangle\leq\|Au\|\|u\|,
\qquad\|Au\|\geq\alpha\|u\|.
$$
Hence $A$ is injective. Its range is closed: if $Au_n$ converges, this last inequality applied to differences makes $u_n$ a <Cauchy sequence>, and its limit maps to the proposed range limit. If $w$ is in the <orthogonal complement> of the range, then $a(u,w)=0$ for all $u$; taking $u=w$ and using coercivity gives $w=0$. The range is thus dense as well as closed, so it is all of $H$. Solve $Au=g$ uniquely; the displayed lower bound gives the asserted estimate.
For complex <Hilbert spaces> the same proof works for a bounded <sesquilinear form>, linear in the first argument, with $\operatorname{Re}a(v,v)\geq\alpha\|v\|^2$. In the convention $a(u,v)=L(v)$, $L$ must then be a bounded conjugate-linear functional represented as $\langle g,v\rangle$.
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