Proving a metric with absolute value I need to prove that function  $\mathbb R × \mathbb R → \mathbb R $ : $f(x,y) = \frac{|x-y|}{1 + |x-y|}$ is a metric on $\mathbb R$. First two axioms are trivial; it's the triangle inequality which is pain. $\frac{|x-y|}{1 + |x-y|}$ + $\frac{|y-z|}{1 + |y-z|} ≥ \frac{|x-z|}{1 + |x-z|} ⇒ \frac{|x-y| + |y-z| + 2|(x-y)(y-z)|}{1 + |x-y| + |y-z| + |(x-y)(y-z)|}≥ \frac{|x-z|}{1 + |x-z|}$, but then I am stuck. Can somebody show me way out of this?
 A: Since $1 + |x-y| + |y -z| \ge 1 + |x-z|$, 
$$1 - \frac{1}{1 + |x-z|} \le 1 - \frac{1}{1 + |x-y| + |y -z|} \\ 
\implies \frac{|x-z|}{1 +|x-z|} \le \frac{|x-y| + |y-z|}{1 + |x-y| + |y-z|} \le \frac{|x-y|}{1 + |x-y|} + \frac{|y-z|}{1 + |y-z|}$$
A: Hint: For $a,b \geq 0$
$$\frac{a+b}{1+a+b} = \frac{a}{1+a+b} + \frac{b}{1+a+b} \leq \frac{a}{1+a}+\frac{b}{1+b}$$
Next, made judicious choices for $a$ and $b$.  
A: Have a look at $g(x)=\frac x{1+x}$. All we need is that for $u,v\ge 0$ and $w\le u+v$, we have $$\tag1 g(w)\le g(u)+g(v).$$
For if this is the case, we can let $u=|x-y|$, $v=|y-z|$, $w=|x-z|$; then $w\le u+v$ by the ordinary triangle inequality, and then $(1)$ i sthe desired triangle inequality for $f$.
To show $(1)$, it is sufficient to show 
$$\tag2 w\le w'\implies g(w)\le g(w')$$
and $$\tag3 g(u+v)\le g(u)+g(v).$$
(Indeed, if $(2)$ and $(3)$ hold, we obtain $(1)$ by letting $w'=u+v$).
We quickly check that $g$ has the following properties:
$$ \tag4 g\text{ is differentiable and }g'(x)\ge 0\text{ for all }x\ge 0$$
which by the MVT implies that $g$ is increasing, i.e. $(2)$.
Moreover we find that $g''(x)\le 0$ for all $x\ge 0$, so that 
$$\tag5 g'(x)\text{ is decreasing}.$$
Therefore 
$$\tag6g(u+v)-g(v)=\int_0^u g'(v+t)\,\mathrm dt\le \int_0^u g'(t)\,\mathrm dt=g(u)-g(0)$$
and using $$\tag7 g(0)=0 $$
this immediately gives us $(3)$.
We conclude that $$f(x,y)=g(|x-y|) $$
obeys the triangle inequality whenever $g\colon[0,\infty)\to\mathbb R$  has the properties $(4)$, $(5)$, and $(7)$.
To ensure that this is a metric, only very little more than this is needed abotu the function $g$, namely that $g(x)=0$ implies $x=0$.
