TITLE: How to show this function is a metric: $d(x,y) = \frac{2|x-y|}{\sqrt{1+x^2}+\sqrt{1+y^2}}$?
QUESTION [5 upvotes]: Consider $( \mathbb{R}, d )$ where $$d(x,y) = \frac{2|x-y|}{\sqrt{1+x^2}+\sqrt{1+y^2}}.$$
The problem arises in showing triangle inequality. I can deal with the numerator only with the help of triangle inequality for the mod in $\mathbb{R}$ but for the denominator I get no clue how to handle?

REPLY [3 votes]: Let $x\geq y\geq z$.
Hence, $$\frac{|x-z|}{\sqrt{1+x^2}+\sqrt{1+z^2}}\geq\frac{|x-y|}{\sqrt{1+x^2}+\sqrt{1+y^2}}$$ it's
$$\frac{x-z}{\sqrt{1+x^2}+\sqrt{1+z^2}}\geq\frac{x-y}{\sqrt{1+x^2}+\sqrt{1+y^2}}$$ or 
$$\frac{y-z}{\sqrt{1+x^2}+\sqrt{1+z^2}}\geq\frac{x-y}{\sqrt{1+x^2}+\sqrt{1+y^2}}-\frac{x-y}{\sqrt{1+x^2}+\sqrt{1+z^2}}$$ or
$$y-z\geq\frac{(x-y)(z^2-y^2)}{\left(\sqrt{1+x^2}+\sqrt{1+y^2}\right)\left(
\sqrt{1+y^2}+\sqrt{1+z^2}\right)}$$ or
$$\left(\sqrt{1+x^2}+\sqrt{1+y^2}\right)\left(
\sqrt{1+y^2}+\sqrt{1+z^2}\right)\geq(y-x)(y+z),$$
which is true because
$$\left(\sqrt{1+x^2}+\sqrt{1+y^2}\right)\left(
\sqrt{1+y^2}+\sqrt{1+z^2}\right)\geq$$
$$\geq(|x|+|y|)(|y|+|z|)\geq|(y-x)(y+z)|\geq(y-x)(y+z).$$
By the same way we can prove that
$$\frac{|x-z|}{\sqrt{1+x^2}+\sqrt{1+z^2}}\geq\frac{|y-z|}{\sqrt{1+y^2}+\sqrt{1+z^2}}.$$
Thus, it's enough to prove that
$$\frac{x-y}{\sqrt{1+x^2}+\sqrt{1+y^2}}+\frac{y-z}{\sqrt{1+y^2}+\sqrt{1+z^2}}\geq\frac{x-z}{\sqrt{1+x^2}+\sqrt{1+z^2}}$$ or
$$\tfrac{x-y}{\sqrt{1+x^2}+\sqrt{1+y^2}}+\tfrac{y-z}{\sqrt{1+y^2}+\sqrt{1+z^2}}\geq\tfrac{x-y}{\sqrt{1+x^2}+\sqrt{1+z^2}}+\tfrac{y-z}{\sqrt{1+x^2}+\sqrt{1+z^2}}$$ or
$$\frac{(x-y)(\sqrt{1+z^2}-\sqrt{1+y^2})}{\sqrt{1+x^2}+\sqrt{1+y^2}}\geq\frac{(y-z)(\sqrt{1+y^2}-\sqrt{1+x^2})}{\sqrt{1+y^2}+\sqrt{1+z^2}}$$ or
$$(x-y)(z^2-y^2)\geq(y-z)(y^2-x^2)$$
$$(x-y)(y-z)(x+y)-(x-y)(y-z)(y+z)\geq0$$ or
$$(x-y)(y-z)(x-z)\geq0,$$ which is obvious.
Done!