The truncated octahedron minimizes surface area among parallelohedra of equal volume

We prove that the regular truncated octahedron uniquely minimizes surface area among all parallelohedra of fixed volume. Equivalently, every three-dimensional parallelohedron $P$ satisfies $$\frac{{\mathcal{H}}^2(\mathrm{\partial }P)}{\mathrm{\mid }P{\mathrm{\mid }}^{2\mathrm{/}3}}\ge \frac{3(1+2\sqrt{3})}{4^{2\mathrm{/}3}},$$ with equality if and only if $P$ is similar to the regular truncated octahedron. Among the non-truncated Fedorov types we prove a stronger sharp bound, attained uniquely by the regular rhombic dodecahedron.