Prove, that $\int_X \nabla u \nabla (v-u) dx + \int_X u(v-u)dx \ge \int_X f(v-u)dx \ \Rightarrow \ -\Delta u + u = f$ where $u = g$ on $\partial X$

Question

We already proved, that

$$a(u,v) = \int_X \nabla u \nabla v dx + \int_X u v dx$$

is a bilinear, continuous, coercive and symmetric form. So wouldn't the solution here simply be to apply the Stampacchia Theorem (as $\phi(u) = \int_X f u dx$), and because $u$ minimizes

$$E(u) = \frac{1}{2} \int_X (|\nabla u|^2 + u^2)dx - \int_X fu dx$$

then by the Dirichlet Principle, it is also the solution of (in this case) $-\Delta u + u = f$ when $u = g$ on the boundary of $X$?

My Profesor told me to take a look into Brezis' book, it looks like my approach should be right (Stampacchia and then Proposition 8.15)?

Or is there another way how this inequality implies that $u$ must be the solution of that differential equation?