Four points $\vec{a}_i$ in $R^3$ are given such that $\vec{a}_1+\vec{a}_2+\vec{a}_3+\vec{a}_4=\vec{0}$ and $|\vec{a}_1|+|\vec{a}_2|+|\vec{a}_3|+|\vec{a}_4|\leq 2$. I want to find the smallest octahedron such that the following 16 points lie inside (or on the surface) of the octahedron:
$\vec{0}$ (1 point)
$\vec{a}_1$, $\vec{a}_2$, $\vec{a}_3$, $\vec{a}_4$ (4 points)
$\vec{a}_1+\vec{a}_2$, $\vec{a}_1+\vec{a}_3$, , $\vec{a}_1+\vec{a}_4$, $\vec{a}_2+\vec{a}_3$, , $\vec{a}_2+\vec{a}_4$, $\vec{a}_3+\vec{a}_4$ (6 points)
$\vec{a}_1+\vec{a}_2+\vec{a}_3$, $\vec{a}_1+\vec{a}_2+\vec{a}_4$, ... (all sums of three vectors) (4 points)
$\vec{a}_1+\vec{a}_2+\vec{a}_3+\vec{a}_4$ (1 point)
Due to the conditions, always two points lie opposite to each other (around the origin). For instance $\vec{a}_1+\vec{a}_2+\vec{a}_3=-\vec{a}_4$, $\vec{a}_1+\vec{a}_3=-(\vec{a}_2+\vec{a}_4)$ and so on. In total, the points form a lattice and I know that an octahedron is the ball of the L1 norm. So the problem is equivalent to finding a suitable coordinate frame in which the Maximum of the L1 norm of these points is minimal. I do not need to know the precise Minima, it would be already enough to show that it is always less than 1 (in case it is true).
Any comments or literature is highly appreciated! :)