Uniform covers theorem ($|K{|}^k\le {\prod }_{A\in \mathcal{A}}|K_A|$)

If $\mathcal{C}$ is a $k$-uniform cover of $[n]$ and $S\subseteq {\mathbb{R}}^n$ is a Euclidean body, thenSlicing in one coordinate, applying the theorem inductively to each slice, and then applying Hölder's inequality to the $k$ projected slice functions proves the inequality.

Exact covering is essential for unrestricted real volumes. Finite subsets of a Cartesian product satisfy the analogous cardinality inequality, and in that discrete version covering at least $k$ times suffices.

A Euclidean body has an axis-parallel box of the same volume with every coordinate-projection volume no larger than its own. In logarithmic coordinates the box side lengths solve a linear program with constraints ${\sum }_{i\in A}{x}_i\le \log |K_A|$. Its dual uses nonnegative exact fractional covers; the uniform cover inequality bounds every dual objective below by $\log |K|$, proving existence by strong duality.

An alternative proof minimizes an array of candidate projection volumes subject to the finitely many inequalities from irreducible uniform covers. Tight constraints force the array to factor into its singleton coordinates, which become the side lengths of the box.

Set $x_i=\log b_i$ for candidate axis-parallel box side lengths. Its coordinate-projection constraints are the displayed linear program. The dual feasible weights are fractional uniform covers. Its optimal extreme point is rational, so the uniform covers theorem forces its objective to be at least $\log |S|$. Weight one on the full coordinate set attains that value. Strong duality therefore supplies box side lengths with volume $|S|$ and no larger coordinate projections.

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.

Equality in the three-dimensional Loomis--Whitney inequality forces a measurable body to agree up to a null set with a Cartesian product $E_1\times E_2\times E_3$. This follows from the equality conditions in the two Cauchy-Schwarz inequalities used in its proof. If the body is connected and is a finite union of positive-volume axis-parallel boxes, each $E_i$ is an interval and equality up to a null set upgrades to exact equality with one box.