TITLE: Does this cross-product norm inequality hold?
QUESTION [6 upvotes]: I asked this on MSE over a month ago, but the one answer I got doesn't seem to work.

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.)

REPLY [6 votes]: The alleged inequality is true. 
Using the definition of the (vector) cross product, for unit vectors $x, y \in R^3$ the original claim boils down to showing that 
\begin{equation*}
  d(x,y) = \sin(\cos^{-1}(x^Ty))
\end{equation*}
is a distance. But it turns out that $d(\cdot,\cdot)$ is actually is a distance on unit vectors in $R^n$.
Showing the above function to be a distance is equivalent to showing that
\begin{equation*}
  \sqrt{1 - (x^Ty)^2}
\end{equation*}
is a distance over the set of unit vectors in $R^n$. But this is a known result of Wang and Zhang, "A trace inequality for unitary matrices", AMM, 1994, pp. 453-455.
EDIT: Also worth comparing with the recent determinant based distance described in this MO question (which suggests that actually after suitable normalization, other symmetric matrix functions should also generate distances, e.g., $d(X,Y) := [1-\text{trace}(\wedge^k(X^TY))]^{1/2}$.