<s>[INST] 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?
 [/INST] 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
$$ \begin{align} D_1+D_2+D_3 &= |a+b-c|+|a+c-b|+|b+c-a| \\
& \ge |a+b-c+a+c-b+b+c-a| = |a+b+c| = D_0 \end{align} $$
</s>