Filling a gap in triangle inequality proof for $d(x,y)= \left\lvert \frac{|x-y|}{1+|x-y|} \right\lvert $.

Question

For the function $d:\mathbb{R} \times \mathbb{R} \to \mathbb{R}$ defined as $$d(x,y)= \left\lvert \frac{|x-y|}{1+|x-y|} \right\lvert. $$ I want to prove this is a metric but Im having issues proving the triangle inequality. I know this problem has been discused before and the one idea im trying to prove this is using the fact that $f(x)=\frac{x}{1+x}$ is increasing and also satisfies the subadditivity property. This is $f(a+b) \leq f(a) + f(b)$ for every $a,b \geq 0$. Im stuck proving the subadditivity, someone mention this can be done with intermediate value theorem but I dont see how. Thanks for reading and helping.

Answer 1

Note also that $f(x) \ge 0$ for $x \ge 0$. $$d(x,z) = f(|x-z|) \le f(|x-y|+|y-z|) \le f(|x-y|) + f(|y-z|) = d(x,y) + d(y,z).$$ The first inequality is due to $f$ being increasing, and the secoond inequality is the subadditivity property.

Answer 2

Note that $f(x)=\frac{x}{1+x}=1-\frac{1}{1+x}$. For $a,b\geq 0 $, $$f(a+b)=1-\frac{1}{1+a+b}.$$ Observe that $1-\frac{1}{1+a+b}\leq 1-\frac{1}{1+a}$(why?). Since $1-\frac{1}{1+b}\geq 0$, the inequality follows directly.