I am trying to prove that if
$$\int_\Omega\frac{|\nabla w|^2}{2} - wf dx$$
Is a functional with a minimizer of $u$ among all possible $w$, then:
$$\int_\Omega \nabla u\cdot\nabla(w - u)dx \geq \int_\Omega (w-u)f dx$$
So far I have managed to prove the following:
Assume $u$ is such a minimum, then by definition we get:
$$\int_\Omega \frac{|\nabla w|^2}{2} - wfdx \geq \int_\Omega \frac{|\nabla u|^2}{2} dx - ufdx$$
But then:
$$\int_\Omega \frac{|\nabla w|^2}{2} -\frac{|\nabla u|^2}{2} dx \geq \int_\Omega wf - ufdx$$
Which gives us:
$$\int_\Omega \frac{(\nabla w - \nabla u) \cdot (\nabla w + \nabla u)}{2} dx \geq \int_\Omega wf - ufdx$$
This is so close, but it's missing just the last little bit. I have tried using the triangle inequality or adding positive terms to the lhs to turn it into $\nabla u \nabla(w - u)$. Could someone throw me a hint for that last step?