Mahler-type volume inequality for convex bodies with tetrahedral symmetry
Arkadiy Aliev
TL;DR
Let $K\subset \mathbb{R}^n$ and $(K-K)^{\circ}$ denote the polar of the difference body; the paper addresses Mahler-type lower bounds for the product $|K| \lvert (K-K)^{\circ}\rvert$ in two regimes. It provides a new proof in the plane ($n=2$) that $|K| \lvert (K-K)^{\circ}\rvert \ge \frac{3}{2}$, with equality for a triangle, using an affine-regular hexagon inscribed in $(K-K)^{\circ}$ and a partition-based area argument. It then treats 3D bodies with tetrahedral symmetry, showing $|K| \lvert (K-K)^{\circ}\rvert \ge \frac{2}{3}$ with equality for tetrahedra, by dissecting $(K-K)^{\circ}$ with fourteen planes and exploiting sectional Mahler bounds via projections. Together, these results yield a specialized, tight validation of a Mahler-type inequality for tetrahedrally symmetric bodies and provide a new 2D method that complements previous proofs. The work reinforces the connection between the volume product, symmetry, and lattice-density considerations in convex geometry.
Abstract
Let $ K $ be a convex body in $ \mathbb{R}^n $. We denote the volume of $ K $ by $ \vert K\vert $, and the polar body of its difference body $ K - K $ by $ (K - K)^{\circ} $. We provide a new proof of the well-known estimate \[ |K||(K - K)^{\circ}| \geq \frac{3}{2} \] for $ K \subset \mathbb{R}^2 $, with equality attained for a triangle. For $ K \subset \mathbb{R}^3 $ with tetrahedral symmetry, we prove that \[ |K| |(K - K)^{\circ}| \geq \frac{2}{3}, \] with equality attained for a tetrahedron.