Diagonals of parallelepiped have length $2,3,5,11$ Is it possible that the four body diagonals (not face diagonals) of a parallelepiped have lengths $2,3,5$, and $11$?
I guess the answer is no, because it is hard for one diagonal to be longer than even the sum of the remaining diagonals ($11>2+3+5$). Maybe there is a corresponding version of the triangle inequality for parallelepiped?
 A: Yes, there is such an inequality. Place the parallelogram with one corner at the origin and the neighboring three corners at $a$, $b$, and $c$. The remaining four corners are at $a+b$, $a+c$, $b+c$ and $a+b+c$.
The length of the body diagonals are
$$ \begin{array}{c} D_0 = |a+b+c| \\ D_1 = |a+b-c| \\ D_2 = |a+c-b| \\ D_3 = |b+c-a| \end{array} $$
Using the triangle equality we get
$$ \begin{align} D_1+D_2+D_3 &= |a+b-c|+|a+c-b|+|b+c-a| \\
& \ge |a+b-c+a+c-b+b+c-a| = |a+b+c| = D_0 \end{align} $$
A: This is a proof for a rectangular parallelepiped only. Let one vertex be the origin, and let the lengths along the three directions be $\mathbf{x, y, z}$. The lengths of the three face diagonals will be ${\mathbf{\sqrt{x^2 + y^2}}}$, $\mathbf{\sqrt{y^2 + z^2}}$ and $\mathbf{\sqrt{z^2 + x^2}}$ and that of the body diagonal = $\mathbf{\sqrt{x^2 + y^2 + z^2}}$. Now it is easy to prove that the sum of the lengths of the face diagonals exceed the length of the body diagonal.
A: Another inequality to go with Henning's is this:
$$3|-a+b+c|^2+3|a-b+c|^2+3|a+b-c|^2-|a+b+c|^2\\=4|a-b|^2+4|a-c|^2+4|b-c|^2\ge0$$
