# Metric $d(x,y)=\frac{|x-y|}{\sqrt{1+x^2}\sqrt{1+y^2}}$ on $\mathbb{R}$

Define the following function on $\mathbb{R}$ by $d(x,y)=\dfrac{|x-y|}{\sqrt{1+x^2}\sqrt{1+y^2}}$. Prove that this is metric.

I proved the first two properties of metric. But how to prove that $d(x,y)+d(y,z)\geqslant d(x,z)$ for any $x,y,z\in \mathbb{R}$

• The inequality sign is incorrect. – user99914 Nov 4 '15 at 10:22
• You've got your triangle inequality backwards, maybe that's why you're having a hard time proving it? ;) – Aaron Golden Nov 4 '15 at 10:23
• Please sorry dear guys! :)) – ZFR Nov 4 '15 at 10:25
• @JohnMa, I edited. Sorry for mistake – ZFR Nov 4 '15 at 10:25
• But I still don't know how to prove it :( Can anyone show a solution? – ZFR Nov 4 '15 at 10:27

Hint: Show that $$d(x,y)=|\sin({\rm Arctan}(x)-{\rm Arctan}(y))|$$ and then use that $|\sin(a+b)|\leq |\sin(a)|+|\sin(b)|$.