How to show that the spherical metric satisfies the triangle inequality? For $x,y\in \mathbb R^n$ define 
$$d(x,y)={\|x-y\| \over \sqrt{1+\|x\|^2} \sqrt{1+\|y\|^2}}$$
Here $\|x\|$ is the euclidean norm of a vector. How to prove that $d$ (the spherical metric) is indeed a metric? 

Progress so far: 


*

*$d(x,y)\ge 0$ is obvious.

*$d(x,y) =0 \iff \|x-y\|=0 \iff x=y$, so the positivity holds.

*$d(x,y) = d(y,x)$ is clear from the formula, so symmetry holds. 


But I am having difficulties with the triangle inequality. Writing it out in coordinates leads to a complicated inequality with square roots all over the place in denominators. Is there a better way? 
 A: The quickest way I know is to "cheat"${}^*$ with Stereographic projection. Introduce the map $F:\mathbb R^n\to\mathbb R^{n+1}$ defined by $F(  x)=( z,t)\in\mathbb R^n\times \mathbb R$ with 
$$  z=\frac{x}{1+\|x\|^2},\quad t= \frac{\|x\|^2}{1+\|x\|^2}$$
(This is a projection onto the sphere $\|z\|^2+(t-1/2)^2=1/4$, but this fact isn't needed.)
Direct computation shows that 
$$
\|F(x)-F(y)\|^2  = \frac{\|x\|^2}{(1+\|x\|^2)^2}+\frac{\|y\|^2}{(1+\|y\|^2)^2} - \frac{2x\cdot y}{(1+\|x\|^2)(1+\|y\|^2)} + \frac{1}{(1+\|x\|^2)^2}+\frac{1}{(1+\|y\|^2)^2} - \frac{2}{(1+\|x\|^2)(1+\|y\|^2)}$$
which simplifies to
$$
\frac{1}{1+\|x\|^2}+\frac{1}{1+\|y\|^2} - \frac{2x\cdot y}{(1+\|x\|^2)(1+\|y\|^2)}  - \frac{2}{(1+\|x\|^2)(1+\|y\|^2)}
$$
and subsequently to
$$
\frac{2+\|x\|^2+\|y\|^2-  2x\cdot y -2}{(1+\|x\|^2)(1+\|y\|^2)}  = {\|x-y\|^2 \over ( 1+\|x\|^2)\,(1+\|y\|^2)}
$$
Thus, $\|F(x)-F(y)\|=d(x,y)$, and the triangle inequality for $d$ follows from the triangle inequality for the Euclidean distance in $\mathbb R^{n+1}$.

$(*)$ I think this is not much cheating, because what use is this metric to us without knowing its relation to the sphere? 
