# Prove $d(x,y)=\left|\frac{x}{1+\sqrt{1+x^2}}-\frac{y}{1+\sqrt{1+y^2}}\right|$ is a metric on $\mathbb{R}$.

I have to prove that $$d(x,y)=\left|\frac{x}{1+\sqrt{1+x^2}}-\frac{y}{1+\sqrt{1+y^2}}\right|$$ is a metric on $$\mathbb{R}$$. I managed to prove the non-negativity, symmetry and triangle inequality, but I am stuck on proving $$d(x,y)=0\Leftrightarrow x=y$$. In my textbook I have the following indication: "prove that $$f(x)=\frac{x}{1+\sqrt{1+x^2}}$$ is a strictly increasing function". How this indication helps me, why do I have to prove that the given function is strictly increasing?

• When you have that a function is strictly increasing it's injective, hence from $$f(x) = f(y)$$ it follows that $$x = y$$. Additionally you have $$d(x,y) = 0 \iff f(x) = f(y)$$ so alltogether you get $$d(x,y) = 0 \iff x=y$$ – Gono Feb 23 '20 at 13:30

Consider $$f(x)=\dfrac{x}{1+\sqrt{1+x^2}}$$. Its derivative is $$f'(x)=\dfrac{1}{\sqrt{1+x^2}(1+\sqrt{1+x^2})}$$ is clearly positive then $$f$$ is increasing and consequently injective. So $$f(x)=f(y)\Rightarrow x=y$$ which proves the not obvious side of the separation axiom $$d(x,y)=0\iff x=y$$. The symetry $$d(x,y)=d(y,x)$$ is clear. It remains the triangular inequality.
We have $$\left|\dfrac{x}{1+\sqrt{1+x^2}}-\dfrac{y}{1+\sqrt{1+y^2}}\right|+\left|\dfrac{y}{1+\sqrt{1+y^2}}-\dfrac{z}{1+\sqrt{1+z^2}}\right|\hspace{10mm}(*)$$ By the triangular inequality in $$(\mathbb R,|\space|)$$ one has that $$(*)$$ is greater or equal than $$\left|\dfrac{x}{1+\sqrt{1+x^2}}-\dfrac{z}{1+\sqrt{1+z^2}}\right|$$ so we have prove that $$d(x,y)+d(y,z)\ge d(x,z)$$
Since $$f(x) = \frac{x}{1+\sqrt{1+x^2}}$$ is strictly increasing, suppose without loss of generality that $$x.
$$\implies f(x) < f(y) \implies d(x,y)=|f(x)-f(y)|>0 \qquad (1)$$
The proof where $$f(x)=f(y) \implies d(x,y) = |f(x)-f(y)|=|f(x)-f(x)|=0$$ is trivial. In the other direction, given that $$|f(x)-f(y)|=0$$, if either $$x or $$y but $$d(x,y)= 0$$ then this presents a contradiction to (1).