<s>[INST] 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?
 [/INST] Since $1 + |x-y| + |y -z| \ge 1 + |x-z|$, 
$$1 - \frac{1}{1 + |x-z|} \le 1 - \frac{1}{1 + |x-y| + |y -z|} \\ 
\implies \frac{|x-z|}{1 +|x-z|} \le \frac{|x-y| + |y-z|}{1 + |x-y| + |y-z|} \le \frac{|x-y|}{1 + |x-y|} + \frac{|y-z|}{1 + |y-z|}$$
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