Show that $(X,d)$ is a metric space where $X =\Bbb C $ and the distance function is defined as:
$$d(x,y) = \frac {2|x-y|}{\sqrt {1+|x|^2} + \sqrt {1 + |y|^2}}, \text{ for } x,y \in \Bbb C.$$
I have done the proof of the first two propositions for being a metric, but I'm having a problem in proving the triangle inequality.
Best Answer
First of all, let's get rid of the square roots and other red herrings. The problem is really to show that $\frac{|x-y|}{|x|+|y|}$ is a distance in $\mathbb R^n\setminus\{0\}$. If you want the square root back, just consider the points on the hyperplane at the distance $1$ from the origin.
Second, let us use Alex Ravsky's observation that the case $|y|\le \min(|x|,|z|)$ is simple (just increase the denominators on the left to $|x|+|z|$) and the case $|y|\ge \max(|x|,|z|)$ can be reduced to the previous one by the inversion $x\mapsto \frac{x}{|x|^2}$ which leaves our distance invariant.
So we assume that $r=|x|<\rho=|y|<R=|z|$. Then $$ |x-z|^2=R^2+r^2-2Rr\cos\alpha=(R-r)^2+Rr(2\sin\frac\alpha 2)^2=(R-r)^2+Rr|x'-z'|^2 $$ where $\alpha$ is the angle between $x$ and $z$ and $x'$ is the unit vector in the direction of $x$. Hence, $$ d(x,z)^2=\left(\frac{R-r}{R+r}\right)^2+\frac{Rr}{(R+r)^2}|x'-z'|^2 $$ Similar expressions can be obtained for $d(x,y$ and $d(y,z)$. Note now that due to the inequality $r\le \rho\le R$, the ratio $A(R,r)=\frac{Rr}{(R+r)^2}$ is smaller than both ratios $A(R,\rho)$ and $A(\rho,r)$, so if we use the common value $A=A(R,r)$ everywhere, we shall get a stronger inequality.
Thus, it suffices to prove that $$ \sqrt{\left(\frac{\rho-r}{\rho+r}\right)^2+A|x'-y'|^2} + \sqrt{\left(\frac{R-\rho}{R+\rho}\right)^2+A|y'-z'|^2} \ge \sqrt{\left(\frac{R-r}{R+r}\right)^2+A|x'-z'|^2}\,. $$ By the Minkowski inequality, it will suffice to show that $$ \frac{\rho-r}{\rho+r}+\frac{R-\rho}{R+\rho}\ge \frac{R-r}{R+r} $$ and $$ |x'-y'|+|y'-z'|\ge |x'-z'|\,. $$ The second inequality is just the triangle inequality in $\mathbb R^n$ and the first one is the combination of the convexity of the function $t\mapsto 1/t, t>0$ and the identity $$ \frac{R-\rho}{R-r}(R+\rho)+\frac{\rho-r}{R-r}(\rho+r)=R+r\,. $$