Is there any simple way of proving triangle inequality for $d(x,y)= \dfrac{2|x-y|}{\sqrt{1+x^2}+\sqrt{1+y^2}}, \,\,x,y\in \mathbb{R}$

Question

$d(x,y)= \dfrac{2|x-y|}{\sqrt{1+x^2}+\sqrt{1+y^2}}, \,\,x,y\in \mathbb{R}$

Following are the approaches I took, but can't think further -

1. $d(x,y)= \dfrac{2|x-y|}{\sqrt{1+x^2}+\sqrt{1+y^2}} = \dfrac{2}{|x+y|}\dfrac{|x^2-y^2|}{\sqrt{1+x^2}+\sqrt{1+y^2}} = \dfrac{2}{|x+y|}\dfrac{|x^2 +1 -(1+y^2)|}{(\sqrt{1+x^2}+\sqrt{1+y^2})} = \dfrac{2}{|x+y|} |\sqrt{1+x^2}-\sqrt{1+y^2}| $

Obviously, since denominator was fully positive. Now, the numerator serves easy for the triangle inequality but I'm not getting how to tackle with $|x+y|$ part in the denominator. Any suggestions?

2. If the metric given was $d(x,y)= \dfrac{2|x-y|}{\sqrt{1+x^2}\sqrt{1+y^2}}$ instead, triangle inequality can be shown using $\sin\left(\tan^{-1}\dfrac{x-y}{1+xy}\right) = \sin(\tan^{-1}x - \tan^{-1}y)=\cdots$

and using inequality $|\sin(a+b)|\leq |\sin a|+|\sin b|$. To push in this thing, I used A.M-G.M inequality to get the following,

$\dfrac{2}{\sqrt{1+x^2}+\sqrt{1+y^2}}|x-y|\leq \dfrac{|x-y|}{\sqrt{(\sqrt{1+x^2}\sqrt{1+y^2})}}$

I almost had the required in the RHS, but the whole square root in denominator spoilt. Can't get what further can I do with this thing.

3. I really have no idea about this, but I just took the above but further to ..- H.M inequality.

$\dfrac{2|x-y|}{\sqrt{1+x^2}+\sqrt{1+y^2}} \leq \dfrac{1}{2}\left(\dfrac{|x-y|}{1+x^2}+\dfrac{|x-y|}{1+y^2}\right)$

Is there any way like to define another metric $d^*$, using the given metric $d$, of form $d^*(x,y)= f(x,y)d(x,y)$ where $f(x,y)=f(y,x)$?

Proving triangle $$f(x,y)d(x,y) \leq f(x,z)d(x,z)+f(z,y)d(z,y)$$ even then could be difficult I suppose though.