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Represent each element of $S_1+\cdots +S_n$ by the lexicographically least tuple with that sum. Every coordinate projection of the resulting tuple set is itself lexicographically least for its partial sum: replacing it by a smaller representative would improve the original tuple without changing its sum. Consequently projected cardinalities are at most the corresponding partial sumset cardinalities. Projection inequalities, including the box theorem, therefore give sumset inequalities.

Use the lexicographic embedding of a sumset: choose the least tuple $(s_1,\dots ,s_n)\in {\prod }_i{S}_i$ representing each distinct sum, and let $B$ be the set of those tuples. Then $|B|=|S|$. If two projections onto $A$ had the same partial sum but different tuples, replacing the larger projected tuple by the smaller one inside its full representative would preserve that full sum and improve its lexicographic order. This is impossible. Thus the summation map is injective on the projection $B_A$, giving $|B_A|\le |S_A|$.

Replace every integer tuple of $B$ by the closed unit cube centered there. Their interiors are disjoint, and each coordinate projection is a union of unit cubes indexed by $B_A$, so the resulting body has volume $|B|$ and projected volume $|B_A|$. Apply the box theorem to obtainThe empty-set case uses $S_{\varnothing }=\{0\}$ and the empty product one.

The uniform cover inequality states that for a positive-volume body $K\subset {\mathbb{R}}^n$ and a multiset $\mathcal{A}$ in which each coordinate occurs exactly $k$ times, $\boxed{|K{|}^k\le \prod _{A\in \mathcal{A}}|K_A|}$, where $K_A$ is its coordinate projection and volumes use the corresponding dimensions.

To deduce the box theorem, maximize ${\sum }_i{x}_i$ subject to ${\sum }_{i\in A}{x}_i\le \log |K_A|$ for every nonempty $A\subseteq [n]$. This linear program is feasible and bounded above by the full-set constraint. Its dual minimizes ${\sum }_A{\lambda }_A\log |K_A|$ over nonnegative weights with ${\sum }_{A\ni i}{\lambda }_A=1$. For rational weights, clearing denominators gives an exact uniform cover, so the uniform cover inequality bounds that dual objective below by $\log |K|$. The same holds for all feasible weights by density of rational points in the rational constraint polyhedron. Strong duality makes the primal optimum at least $\log |K|$, while the full-set constraint makes it at most that value. An optimal solution therefore gives $b_i=e^{x_i}>0$ withThe axis-aligned box with these side lengths has the required volume and projection volumes.