= Solution
This is the <Dirichlet principle>. Put $v=w-\psi$, so $v$ vanishes on the boundary. For real, sufficiently regular functions of finite <Dirichlet energy>, <Green's first identity> gives
$$
\int_V\nabla\psi\cdot\nabla v\,dV=\int_Sv\,\partial_n\psi\,dS-\int_Vv\nabla^2\psi\,dV=0.
$$
The surface term is zero because $v=0$, and the volume term is zero because $\psi$ is a <harmonic function>. Expanding the square of $\nabla w=\nabla\psi+\nabla v$ now gives the stronger identity
$$
\boxed{\int_V|\nabla w|^2\,dV=\int_V|\nabla\psi|^2\,dV+\int_V|\nabla(w-\psi)|^2\,dV\ \geq\ \int_V|\nabla\psi|^2\,dV.}
$$
Thus the <harmonic function> minimizes <Dirichlet energy> among functions with the same boundary values. Equality requires $\nabla v=0$, hence $v$ is constant on each connected component; the zero boundary value makes that constant zero on every component having boundary. For an unbounded domain one additionally needs the boundary term at infinity to vanish; the usual bounded regular-domain setting needs no such extra condition.
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