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?
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?