The truncated octahedron minimizes surface area among parallelohedra of equal volume

Abstract

We prove that the regular truncated octahedron uniquely minimizes surface area among all parallelohedra of fixed volume. Equivalently, every three-dimensional parallelohedron $P$ satisfies \[ \frac{\mathcal H^2(\partial P)}{|P|^{2/3}} \ge \frac{3(1+2\sqrt3)}{4^{2/3}}, \] 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.

arXiv commentChatGPT (GPT-5.6 Sol) was used during the preparation of this work to check algebraic identities and to assist in exploring and verifying proof arguments. In the second version corrected some misprints, and moreover reorganized some parts (in particular relative to the entropy contraction estimates) to improve readibility