Evans stated in his book about PDEs that the Lax-Milgram can be applied to $B_{\mu}$, but I can not see how because is not clear for me how $B_{\mu}$ is coercive. I would like to know how $B_{\mu}$ is coercive.
Thanks in advance!
$\textbf{EDIT:}$
I apologize for not include previously some relevant informations, but I included below now.
I hope this is what your were looking for: The bilinear form $B_{\mu}[u,v]=B[u,v]+\mu(u,v)$ associated with the operator $L$ satisfies the coercivity conditions of the Lax-Milgram-Lemma because $$B_{\mu}[u,u]=B[u,u]+\mu(u,u)\ge\beta||u||_{H^{1}}^2+(\mu-\gamma)(u,u) = \beta||u||_{H^{1}}^2+(\mu-\gamma)||u||_{L^{2}} \ge \beta ||u||_{H^{1}}$$ We used that $\mu\ge\gamma$ (where $\gamma$ was derived in the energy estimates of the theorem you included) and that $(u,u)$ is the $L^{2}$ inner product