Elliptic equations and variational formulation

Question

My question: We now consider the following equation: $$ \left\{\begin{array}{l} -\Delta_x u+u(x)=f(x), \quad x \in \Omega \\ \left\langle\nabla_x u(x), \vec{n}(x)\right\rangle=g(x), \quad x \in \partial \Omega . \end{array}\right. $$ Prove that solutions of this equation satisfy the following variational equality: for any $v \in C^1(\mathbb{R})$, $$ \int_{\Omega}\left(\nabla_x u(x) \cdot \nabla_x v(x)+u(x) v(x)\right) d x=\int_{\Omega} f(x) v(x) d x+\int_{\partial \Omega} g(x) v(x) d s(x) . $$ Then, prove that any $u \in C^2(\bar{\Omega})$ satisfying the variational equality satisfies the equation above.
My attempt Let's recall the integration by part formula $u, w \in C^1(\bar{\Omega})$ and $i \in\{1, \ldots, N\}$ $$ \int_{\Omega} u(x) \frac{\partial w}{\partial x_i}(x) d x=-\int_{\Omega} w(x) \frac{\partial u}{\partial x_i}(x) d x+\int_{\partial \Omega} u(x) w(x) d s(x) $$ Using this formula with $w=\frac{\partial v}{\partial x_i}(x)$, i can prove that $$\int_{\Omega} \Delta_x u(x) v(x) d x=-\int_{\Omega} \nabla_x u(x) \cdot \nabla_x v(x) d x+\int_{\partial \Omega}\langle\nabla u(x), \vec{n}(x)\rangle v(x) d s(x)$$ Besides, we have $\Delta_x u=u(x) - f(x)$. That implies $$\int_{\Omega}\left(\nabla_x u(x) \cdot \nabla_x v(x)+u(x) v(x)\right) d x=\int_{\Omega} f(x) v(x) d x+\int_{\partial \Omega} g(x) v(x) d s(x).$$ But i don't know how to deal with the next question: prove that any $u \in C^2(\bar{\Omega})$ satisfying the variational equality satisfies the equation above.