# 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:

1. $d(x,y)\ge 0$ is obvious.
2. $d(x,y) =0 \iff \|x-y\|=0 \iff x=y$, so the positivity holds.
3. $d(x,y) = d(y,x)$ 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?

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Cauchy Schwartz inequality ? – Maman Jan 3 '15 at 23:35
Haven't tried that, how would it look?does it work? – Bozo Vulicevic Jan 3 '15 at 23:42
@BozoVulicevic Why don't you look it up before asking such a question. Don't say "does it work?" because it sounds like you are expecting everyone to answer your question as a response to nagging, without doing work yourself. Try it and see. – Suzu Hirose Jan 4 '15 at 0:08
@Suzu Hirose I apologize for not being honest, i have tried that to no avail, like many other techniques as well. None have worked and i apologize again for not displaying my efforts for vividly for the rest to see. What does not work is not worth mentioning i believe. A lack of effort on my part is not present. – Bozo Vulicevic Jan 4 '15 at 0:44
I would use stereographic projection: introduce 3D coordinates $u,v,w$ related to $x,y$ by the projection formulas; check that the distance function you have corresponds to the usual Euclidean distance in 3D space; conclude. – ᴡᴏʀᴅs Jan 4 '15 at 1:01

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?