Gradient of harmonic function matching with $C^2$ function on boundary is bounded by gradient of function Let $\Omega \subset \mathbb{R}^n$, $n\geq 1$ be a bounded domain with $\partial \Omega \in C^1$ and $u \in C^2(\Omega)\cap C^1(\overline\Omega)$ a harmonic function.
Show that for all $v\in C^2(\Omega)\cap C^1(\overline\Omega)$ with $v=u$ on $\partial \Omega$ the following holds:
$$ \int_\Omega |\nabla u|^2 dx\leq \int_\Omega |\nabla v|^2dx  $$
I got the hint that $0\leq \int_\Omega |\nabla (u-v)|^2 dx$ but I simply couldn't come up with anything useful from that.

i)
\begin{align*} 0&
=\int_{\partial \Omega } \ (u-v)\nabla (u-v)\cdot n
\\&=_{divergence\ theorem} \int \ {\rm div}\ (u-v)\nabla (u-v)
\\&=\int\ |\nabla (u-v)|^2+(u-v)\Delta (-v) \end{align*}
Hence $$ A:=\int\ u\Delta v -v\Delta v \geq 0 $$
ii) $(u-v)\nabla v$ : $$ 0= \int_{\partial \Omega }\ n\cdot
(u-v)\nabla v = \int\ {\rm div}\ (u-v)\nabla v =
 \int \ \nabla u\cdot
\nabla v-|\nabla v|^2 +A$$ so that
$$\int\ |\nabla v|^2\geq
\int\ \nabla u\cdot \nabla v $$
iii) $v\nabla u$ : Here $$ \int_{\partial\Omega }\ n\cdot
u\nabla u= \int\ {\rm div}\ v\nabla u = \int\ \nabla v\cdot \nabla u
$$
And $$ \int_{\partial \Omega }\ n\cdot u \nabla u = \int\ |\nabla
u|^2$$
