Let $G\subset \mathbb{R}^n$ be an open set with smooth boundary and let $u,v:\bar{G}\rightarrow\mathbb{R}$ be $C^2$ functions. Suppose that $u$ is harmonic, i.e. $\Delta u=0$, and that $u|_{\partial G}=v|_{\partial G}$.
Prove that:
1. $$\int_G |\nabla u|^2=\int_G \left \langle \nabla u, \nabla v \right \rangle$$
2. $$\int_G |\nabla u|^2\leq\int_G |\nabla v|^2$$
I managed to solve the first part. It's a simple application of the assumptions and Green's identities, more specifically the second one (which can be found at the very beginning of these notes).
However, I'm stuck on the second one.
Using the first part it suffices to show that
$$\int_G \left \langle \nabla v, \nabla (v-u) \right \rangle\geq 0$$
but I cannot proceed from here.
I'd be glad for someone's help on this one.
We see that if \begin{equation*} \int_{G}\left\lvert\nabla u(x)\right\rvert^{2}dx=0 \end{equation*} then the second inequality is clear. So we may assume this integral is positive. By the equality obtained before we have \begin{align*} \int_{G}\left\lvert\nabla u(x)\right\rvert^{2}dx&=\int_{G}\left<\nabla u(x),\nabla v(x)\right>dx\le\int_{G}\left\lvert \nabla u(x)\right\rvert\left\lvert\nabla v(x)\right\rvert{}dx\\ &\le\left(\int_{G}\left\lvert\nabla u(x)\right\rvert^{2}dx\right)^{\frac{1}{2}}\left(\int_{G}\left\lvert\nabla v(x)\right\rvert^{2}dx\right)^{\frac{1}{2}} \end{align*} Dividing through by $\left(\int_{G}\left\lvert\nabla u(x)\right\rvert^{2}dx\right)^{\frac{1}{2}}$ and squaring gives the desired inequality.