How to show this function is a metric: $d(x,y) = \frac{2|x-y|}{\sqrt{1+x^2}+\sqrt{1+y^2}}$?

Question

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?

Answer 1

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!

Answer 2

I am going to give you some hints. Let $x,y\in\mathbb{R}$ be fixed. You want to show $d(x,z) + d(z,y) \geq d(x,y)$ for all $z$. If $x=y$ it is trivially true. So without loss of generality assume, $x < y$. Define $f:\mathbb{R}\to\mathbb{R}$ as $f(z) = d(x,z) + d(z,y)$. Note that $f(x) = f(y)$. Try to show that this function is decreasing in the interval $(-\infty,x]$ and increasing in $[y,\infty)$. So $f(z) \geq f(x)=f(y)$ for all $ z\in (-\infty,x]\cup[y,\infty)$. In the interval $[x,y]$ $f$ is smooth with $f^{\prime\prime} < 0$ i.e $f$ is concave. So $f(z) \geq f(x)$ in $[x,y]$ also.