Loomis--Whitney inequality

The sets $[n]\setminus \{i\}$ form an $(n-1)$-uniform cover, so the uniform covers theorem gives

Each nonnegative integrable $f_j$ depends on all coordinates except the $j$th. In dimension three, Cauchy-Schwarz inequality first in the third coordinate and then in the remaining pair proves the displayed estimate. Induction integrates the last coordinate by Hölder's inequality on the first $d-1$ factors, then applies Hölder's inequality to the remaining factor and the dimension-$(d-1)$ product. Taking indicators of coordinate projections of a Euclidean body gives the volume version of the Loomis--Whitney inequality.

Positive numbers $a,b,c$ are the areas of the three coordinate projections of a volume-one Euclidean body exactly when $abc\ge 1$. Necessity is the three-dimensional Loomis--Whitney inequality. For sufficiency take box lengths $x=\sqrt{ac/b}$, $y=\sqrt{ab/c}$ and $z=\sqrt{bc/a}$, and retain the union of the three slabs with first, second or third normalized coordinate at most $t$. All three projections are full rectangles, while volume is $\sqrt{abc}[1-(1-t{)}^3]$. Choosing $t=1-(1-1/\sqrt{abc}{)}^{1/3}$ gives volume one. The slabs have positive thickness, so the construction uses a body with interior rather than zero-volume additions.

For a Euclidean body $S\subseteq {\mathbb{R}}^3$, equalityin the Loomis--Whitney inequality forces $S$ to agree outside a null set with a Cartesian product $E_1\times E_2\times E_3$ of measurable sets. The equality conditions in the two Cauchy-Schwarz inequalities used in the proof give this product structure. Agreement outside a null set alone cannot specify which boundary points belong to $S$.

If $S$ is connected and a finite union of closed, positive-volume axis-parallel boxes, equality does imply that $S$ is exactly one closed axis-parallel box. Each coordinate projection of a Euclidean body onto one axis is then an interval. On the finite grid of box endpoints, the product structure forces every open cell inside the bounding axis-parallel box to belong to $S$. Since $S$ is a closed set and is the closure of its interior, it contains the whole bounding axis-parallel box.

Closedness matters. ConsiderThis connected set is a union of two positive-volume axis-parallel boxes. Its volume and all three projected areas are one, so equality holds. But only part of its top face is present, and it is not a Cartesian product of intervals.