= Solution
<Green's first identity> and the boundary condition give
$$
\int_V|\nabla u|^2dV
=\int_Su\frac{\partial u}{\partial n}dS-\int_Vu\nabla^2u\,dV=0.
$$
Thus $\nabla u=0$, so $u$ is constant on each connected component of $V$; its zero boundary value forces $\boxed{u=0}$.
Since $w-v=0$ on $S$ and $\nabla^2v=0$, another application of <Green's first identity> gives
$$
\int_V\nabla v\mathbin\cdot\nabla(w-v)dV=0.
$$
Consequently
$$
\boxed{\int_V\nabla v\mathbin\cdot\nabla w\,dV
=\int_V|\nabla v|^2dV}.
$$
Expanding $\nabla w=\nabla v+\nabla(w-v)$ and using the vanishing cross term yields the <Dirichlet principle>
$$
\boxed{\int_V|\nabla w|^2dV
=\int_V\left(|\nabla v|^2+|\nabla(w-v)|^2\right)dV
\geq\int_V|\nabla v|^2dV}.
$$
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