Let $P$ be the $1$-skeleton of a convex polyhedron fixed in $\mathbb{R}^3$, and $|P|$ the sum of the Euclidean lengths of the edges of $P$. Let $P_1, P_2, P_3$ be the perpendicular projections of $P$ onto the Cartesian coordinate planes, and $|P_i|$ the sum of the lengths of the segments of $P_i$.

For example, for the particular placement of $P$ a unit edge-length regular tetrahedron shown below, $|P_1|+|P_2|+|P_3|$ is nearly double $|P|=6$:

^{ $|P_1|$ (red) $=1+\sqrt{\frac{2}{3}}+\sqrt{\frac{11}{3}}$. $|P_2|$ (green) $=1+\sqrt{3}$. $|P_3|$ (blue) $=3+\sqrt{3}$. $\Sigma \approx 11.2$. }

Conjecture. For any placement of any convex polyhedron $P$, $|P_1|+|P_2|+|P_3| \ge |P|$, with equality uniquely achieved by the cube.

For a unit edge-length cube $P$, $|P|=12$ and $|P_i|=4$ when oriented so that each projection is a square. So I'm conjecturing that the cube hides its edges in projection more effectively than any other convex polyhedron. Can anyone see a proof or a counterexample?

I would also be interested in which orientations of the regular tetrahedron minimize $\Sigma |P_i|$.

The higher-dimensional analog could be the subject of a future post.