Proving that the triangle inequality holds for a metric on $\mathbb{C}$ 
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.
 A: 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
$$
\frac{R-\rho}{R-r}(R+\rho)+\frac{\rho-r}{R-r}(\rho+r)=R+r\,.
$$
A: A partial result. Let $x,y,z\in\mathbb{C}$ such that $|y|\leq\min\{|x|,|z|\}$ then 
$$d(x,z)=\frac{2|x-z|}{\sqrt{1+|x|^2}+\sqrt{1+|z|^2}}\leq\frac{2|x-y|+2|y-z|}{\sqrt{1+|x|^2}+\sqrt{1+|z|^2}}$$$$=\frac{2|x-y|}{\sqrt{1+|x|^2}+\sqrt{1+|z|^2}}+\frac{2|y-z|}{\sqrt{1+|x|^2}+\sqrt{1+|z|^2}}$$
(Note that the last result is obtained by the triangle inequality on the metric $d^*(x,y)=|x-y|$)
Since $|y|\leq\min\{|x|,|z|\}$ then
$$\sqrt{1+|x|^2}+\sqrt{1+|y|^2}\leq\sqrt{1+|x|^2}+\sqrt{1+|z|^2}$$ 
and
$$\sqrt{1+|y|^2}+\sqrt{1+|z|^2}\leq\sqrt{1+|x|^2}+\sqrt{1+|z|^2}$$
which together with the result above on the main metric $d(x,z)$ yield
$$\frac{2|x-y|}{\sqrt{1+|x|^2}+\sqrt{1+|z|^2}}+\frac{2|y-z|}{\sqrt{1+|x|^2}+\sqrt{1+|z|^2}}$$$$\leq\frac{2|x-y|}{\sqrt{1+|x|^2}+\sqrt{1+|y|^2}}+\frac{2|y-z|}{\sqrt{1+|y|^2}+\sqrt{1+|z|^2}}=d(x,y)+d(y,z)$$
In other words
$$d(x,z)\leq d(x,y)+d(y,z)$$
A: I did a small step before I stuck, but it may be useful for others. So, it seems the following.
The cases $|y|\le\min\{|x|,|z|\}$ and $|y|\ge\max\{|x|,|z|\}$ are simple and already known to us (see Arian’s answer and here; the second case is reduced to the first by substitution $u=x^{-1}$, $v=y^{-1}$, and $w=z^{-1}$). 
So it suffices to consider a case $| x | < | y | < | z |.$ Fix $x$, $z$ and $|y|$. Answering a question for which value of $y$ the left part of the inequality attains minimum, I use geometric optics. 
Let there be light. Let the speed of light in the disk with radius $|y|$ centered at the origin $o$ be $v_1=\sqrt {1+|x|^2} +  \sqrt {1 + |y|^2}$, and $v_2=\sqrt {1+|z|^2} +  \sqrt {1 + |y|^2}$ outside the disk. Fermat's principle states that the light travels the path which takes the least time. From Fermat principle may be derived known Snell–Descartes law of refraction, which yeilds 
$$\frac{\cos\angle oyx}{-\cos\angle zyo}=\frac {v_1}{v_2}.$$ 
But $$\cos\angle oyx=\frac{(y-x,y)}{|y||y-x|},$$
$$- \cos\angle zyo=\frac{(y-z,y)}{|y||y-z|}.$$
So $$\frac{(y-x,y)}{|y-x|\left(\sqrt {1+|y|^2} +  \sqrt {1 + |x|^2}\right)}=
\frac{(y-z,y)}{|y-z|\left(\sqrt {1+|y|^2} +  \sqrt {1 + |z|^2}\right)}.$$
