$u,v$ are scalar fields on $V\subset\mathbb{R}^3$ such that $\nabla^2 u=0$ on $V$ and $u=v$ on $\partial V$. Prove that:
$$\int_V|\boldsymbol{\nabla} v|^2\,\mathrm{d}V\geq\int_V|\boldsymbol{\nabla }u|^2\,\mathrm{d}V$$
I am lost with this problem: could someone offer a hint? It looks like the divergence theorem might come in handy, but I am not sure how.
Denote for convenience $f = v - u$. Then $$ \int_V |\nabla v|^2 dV = \int_V |\nabla u|^2 dV + \int_V |\nabla f|^2 dV + 2 \int_V (\nabla u, \nabla f) dV. $$
Now get rid of the last summand using the fact that $\nabla^2 u = 0$, the divergence theorem, and the fact that $f|_{\partial V} = 0$.