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The problem is to establish the triangle inequality in the extended complex plane $C_\infty$ for the following function:

$d(z,z')=\frac{2|z-z'|}{\sqrt{(1+|z|^2)(1+|z'|^2)}}$; $d(z,\infty)=d(\infty,z)=\frac{2}{\sqrt{(1+|z|^2)}}$; $d(\infty,\infty)=0$.

To prove this I start by letting $x,y,z\in C$. We wish to prove $d(x,y)\le d(x,z)+d(y,z)$. The right hand side of the triangle inequality is now

$$\frac{2|x-z|}{\sqrt{(1+|x|^2)(1+|z|^2)}}+\frac{2|y-z|}{\sqrt{(1+|y|^2)(1+|z|^2)}}$$ which is at least

$$\frac{2(|x-z|+|y-z|)}{\sqrt{(1+|x|^2)(1+|y|^2)(1+|z|^2)}}$$

What to do next?

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    See for example https://math.stackexchange.com/q/844759/42969. – Martin R Oct 21 '21 at 07:15
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    Or this: https://math.stackexchange.com/q/1439664/42969 – Martin R Oct 21 '21 at 07:18
  • @Martin R The validated answer in the first reference you have given is interesting ; the only thing which is absent is the fact that it is a "distance transportation" via inversion (the geometric transform). – Jean Marie Oct 21 '21 at 08:06
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    @JeanMarie: I know that there is another answer which uses the fact that the $d(z, w)$ is exactly the Euclidean distance of the projections of $z, w$ on the Riemann sphere (which is the most natural answer IMO). Unfortunately I haven't found that again yet. – Martin R Oct 21 '21 at 08:09
  • @JeanMarie: Here it is: https://math.stackexchange.com/q/1090222/42969 (found in my bookmarks :) – Martin R Oct 21 '21 at 08:11
  • @Martin R Thank you ! – Jean Marie Oct 21 '21 at 08:15

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