Let $\times$ denote the cross-product. $\;$ Is it the case that
For all unit vectors $\:\mathbf{x}\hspace{.01 in},\hspace{-0.03 in}\mathbf{y}\hspace{-0.03 in},\hspace{-0.02 in}\mathbf{z}\:$ in $\mathbf{R}^3$, $\;\;\;\; \left|\left|\hspace{.03 in}\mathbf{x} \hspace{-0.03 in}\times \hspace{-0.03 in}\mathbf{z}\hspace{.02 in}\right|\right| \:\: \leq \:\: \left|\left|\hspace{.03 in}\mathbf{x} \hspace{-0.03 in}\times \hspace{-0.03 in}\mathbf{y}\hspace{.02 in}\right|\right| \hspace{.02 in}+\hspace{.02 in}\left|\left|\hspace{.03 in}\mathbf{y} \hspace{-0.03 in}\times \hspace{-0.03 in}\mathbf{z}\hspace{.02 in}\right|\right| \;\;\;\;\;$.
?
(If yes, then $\;\;\; \langle [\mathbf{x}]\hspace{.01 in},\hspace{-0.03 in}[\hspace{.02 in}\mathbf{y}] \rangle \: \mapsto \: \left|\left|\hspace{.03 in}\mathbf{x} \hspace{-0.03 in}\times \hspace{-0.03 in}\mathbf{y}\hspace{.02 in}\right|\right| \;\;\;$ defines a nice metric on the projective plane.)
Since they are all unit vectors, the cross product of any two is just the sin of the angle between the two, and is also directly proportional to the area of the triangle formed using the vectors as two of the sides. Therefore, it is equivalent to show that given a tetrahedron ABCD with ||AB|| = ||AC|| = ||AD|| = 1, [ABC] <= [ABD] + [ACD]. Now, there exists a line through A such that the projection of A onto a point A' in the plane containing BCD has ||A'B|| = ||A'C|| = ||A'D||. This projection also maintains the proportions between the areas of the triangles, and so it suffices to show [A'BC] <= [A'BD] + [A'CD]. Moreover, since the range of the cross product is [0, pi], angles BA'C, CA'D, and DA'B also fit into that range.
Now, let a = BA'D, b = CA'D, and so BA'C = 2*pi - a - b, and converting back from triangles to angles, it suffices to show sin(2*pi - a - b) <= sin(a) + sin(b), and thus only that 0 <= sin(a) + sin(b) + sin(a + b), 0 <= a, b <= pi. Moreover, sin(a) + sin(b) + sin(a + b) = cos(a/2)*cos(b/2)*sin((a+b)/2). Finally, since cos(x) is nonnegative for x in [0, pi/2], as is sin(x) for x in [0, pi], the inequality holds and we are done.