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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.

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.

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 uniform covers theorem says that if a multiset $\mathcal{C}$ of coordinate subsets covers every coordinate exactly $k$ times, then a Euclidean body $S\subseteq {\mathbb{R}}^n$ satisfieswhere $S_A$ is its coordinate projection of a Euclidean body. Work with bounded Borel bodies, whose projections are Lebesgue measurable sets; an empty-coordinate projection has measure one.

Prove the theorem by induction on dimension, with the one-dimensional case immediate. Slice at the last coordinate, writing $S_t\subseteq {\mathbb{R}}^{n-1}$. The subsets $A\setminus \{n\}$ still cover each remaining coordinate $k$ times, so induction givesFor $A$ not containing $n$, the projected slice is contained in $S_A$, giving a constant bound. For $A$ containing $n$, its measure is the slice measure $f_A(t)$ of $S_A$. Exactly $k$ members of $\mathcal{C}$ contain $n$, counted with multiplicity. Integrate and apply Hölder's inequality to those $k$ factors:Fubini's theorem identifies $\int f_A=|S_A|$, completing the proof. Exact covering is needed for real volumes, which may be smaller than one.

The Bollobas--Thomason box theorem states that there is an axis-parallel box $B$ with $|B|=|S|$ and $|B_A|\le |S_A|$ for every nonempty coordinate subset $A$. Use the logarithmic linear program for the box theorem: if its side lengths are $b_i=e^{x_i}$, maximize ${\sum }_i{x}_i$ subject toThe linear program is feasible by choosing all $x_i$ sufficiently negative and is bounded above by the full-set constraint. Its dual minimizes ${\sum }_A{\lambda }_A\log |S_A|$ subject to ${\lambda }_A\ge 0$ and ${\sum }_{A\ni i}{\lambda }_A=1$ for each $i$: precisely a fractional uniform cover.

The dual feasible polytope is nonempty, bounded and defined by integer coefficients, so an optimal extreme point has rational coordinates. Multiply its weights by a common denominator $k$. The resulting integer uniform cover and the uniform covers theorem giveThus the dual optimum is at least $\log |S|$; weight one on $A=[n]$ attains equality. Strong duality gives a primal optimum ${\sum }_i{x}_i=\log |S|$. The corresponding box has the required volume and projected-volume bounds, proving the theorem.

For the Loomis–Whitney theorem, take the box supplied by the Bollobas--Thomason box theorem. Each side length occurs in exactly $n-1$ projections omitting one coordinate, soThis is the Loomis--Whitney inequality.

For the three projection areas of a volume-one body, necessity is $1=|S{|}^2\le abc$. Conversely suppose $abc\ge 1$ and setThe box $[0,x]\times [0,y]\times [0,z]$ has projected areas $a,b,c$ and volume $V=\sqrt{abc}\ge 1$. Inside it retain the three slabsFor $t>0$, each projection is the full corresponding rectangle, since the slab in the omitted coordinate supplies every point of that projection. The volume is $V[1-(1-t{)}^3]$. Chooseto make the volume one. This is a compact connected union of three positive-thickness boxes. Hence the complete answer is

A multiset $\mathcal{C}$ of subsets of $[n]$ is a uniform cover of multiplicity $k$ when each coordinate occurs in exactly $k$ members. The uniform covers theorem states that every Euclidean body $S\subseteq {\mathbb{R}}^n$ satisfieswhere $S_A$ is the coordinate projection of a Euclidean body onto the coordinates in $A$.

We prove it by induction on $n$. Split the cover into ${\mathcal{C}}^{-}$, whose members omit $n$, and ${\mathcal{C}}^{+}$, whose members contain $n$. Exactly $k$ members lie in ${\mathcal{C}}^{+}$. For a last-coordinate value $x$, let $S(x)\subseteq {\mathbb{R}}^{n-1}$ be the corresponding slice. Removing $n$ from the members of ${\mathcal{C}}^{+}$ and retaining the members of ${\mathcal{C}}^{-}$ gives a $k$-uniform cover of $[n-1]$. The inductive hypothesis and Fubini's theorem giveApply Hölder's inequality to the $k$ factors in the integral. Sincewe obtain $|S|\le {\prod }_{A\in \mathcal{C}}|S_A{|}^{1/k}$, which is the result after taking the $k$th power. The one-dimensional base case is immediate.

The Bollobas--Thomason box theorem states that for every body $S\subseteq {\mathbb{R}}^n$ there is an axis-parallel box $B$ such thatAn irreducible uniform cover cannot be decomposed into two smaller uniform covers. There are only finitely many such covers of $[n]$: encode a cover by its multiplicity vector in ${\mathbb{N}}^{2^n}$ and apply the Dickson lemma.

Choose a componentwise minimal positive array $(x_A{)}_{A\subseteq [n]}$ satisfyingfor every irreducible $k$-uniform cover $\mathcal{C}$, together withThe actual projection volumes are feasible by the uniform covers theorem, and finiteness gives a minimal array. Every uniform cover is a disjoint union of irreducible ones, so its cover inequality also holds for this array.

Minimality implies that, for each coordinate $i$, some tight uniform-cover inequality can be chosen whose cover contains the singleton $\{i\}$. Indeed, either such an inequality already blocks decreasing $x_{\{i\}}$, or a tight product inequality $x_A={\prod }_{j\in A}{x}_{\{j\}}$ does; in the latter case take a tight cover containing $A$ and replace that occurrence of $A$ by its singleton coordinates. Let these tight covers be ${\mathcal{C}}_i$, of multiplicities $k_i$, and let $K={\sum }_i{k}_i$. Their multiset union is a $K$-uniform cover. Removing one copy of every singleton leaves a $(K-1)$-uniform cover, so comparison of its cover inequality with the product of all the tight equalities yieldsThe singleton cover gives the reverse inequality, henceFor any $A\subseteq [n]$, the one-uniform cover consisting of $A$ and the singletons $\{i\}$ for $i\notin A$ now gives $x_A\ge {\prod }_{i\in A}{x}_{\{i\}}$. The defining product inequality gives the reverse bound. Thus all these quantities are equal. Taking the side lengths of $B$ to be $x_{\{1\}},\dots ,x_{\{n\}}$ proves the theorem.

Finally suppose that the proper body $S\subseteq {\mathbb{R}}^3$ satisfiesThis is equality in the three-dimensional Loomis--Whitney inequality. In its proof, equality must hold in both applications of Cauchy-Schwarz inequality. Their equality conditions force the three projection indicators to factor through one-dimensional measurable sets $E_1,E_2,E_3$, and forceup to a set of Lebesgue measure zero; this is Equality in the three-dimensional Loomis--Whitney inequality.

Because $S$ is connected, each one-coordinate projection is connected and hence is an interval. Because $S$ is a finite union of positive-volume axis-parallel boxes, a proper difference between $S$ and the product of those three intervals would contain a positive-volume rectangular cell in a common finite subdivision. That would contradict equality up to measure zero. Consequently the equality is exact and