= Solution
The identity follows immediately from the product rule:
$$
\nabla\cdot(\kappa\psi\nabla\phi)
=\psi\nabla\cdot(\kappa\nabla\phi)+\kappa\nabla\psi\cdot\nabla\phi.
$$
If $\phi_1,\phi_2$ have the same boundary data, put $\psi=\phi_1-\phi_2$. Integrating the identity with $\phi=\psi$ gives
$$
\int_V\kappa|\nabla\psi|^2dV=0
$$
because the volume equation and boundary term vanish. Positivity of $\kappa$ makes $\psi$ constant, and its boundary value makes it zero, proving uniqueness.
For any $w=\phi+\psi$ with $\psi=0$ on the boundary, expansion gives
$$
\int\kappa|\nabla w|^2
=\int\kappa|\nabla\phi|^2+int\kappa|\nabla\psi|^2,
$$
because the cross term vanishes by the same integration by parts. This proves the inequality and the <Dirichlet principle>.
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