Prove that $|x-y||z| = |y-z||x| + |z-x||y|$ holds iff $x,y,z,0$ are contained on a circle such that the pairs $x,y$ and $z,0$ seperate each other

Question

In the book of Linear Algebra by Werner Greub, at page 191 Q.3, it is asked that,

Prove that $|x-y||z| = |y-z||x| + |z-x||y|$ holds iff $x,y,z,0$ are contained on a circle such that the pairs $x,y$ and $z,0$ seperate each other

I basically couldn't do much thing but the question is really interesting, and I'm sure that there will be good answers both geometrically and axiomatically.

Edit:

Note that this is a direct result of Ptolemy-inequality, but still can't see the result.

Edit2:

Our vector space is a real inner product space.

Answer 1

This means that in the complex plane, $x^{-1}$, $y^{-1}$ and $z^{-1}$ are collinear with $z^{-1}$ between $x^{-1}$ and $y^{-1}$. Therefore $$|x^{-1}-y^{-1}|=|x^{-1}-z^{-1}|+|z^{-1}-y^{-1}|.$$