Proving inequality with absolute value

Question

How can I show the following inequality for any real numbers $x,y,z$?

$$\frac{|x-z|}{1+|x-z|}\le \frac{|x-y|}{1+|x-y|} + \frac{|y-z|}{1+|y-z|}.$$

The triangle inequality could be useful, but I am not sure in what way.

Thank you

Answer 1

Notice that for all $x \in \mathbb{R}$ we have $$ \frac{x}{1+x} = 1 - \frac{1}{1 + x}. $$ So by using the triangle inequality we have \begin{align*} \frac{|x-z|}{1 + |x-z|} &\leq 1 - \frac{1}{1 + |x-z|} \leq 1 - \frac{1}{1 + |x-y| + |y-z|} \\ &= \frac{|x-y| + |y-z|}{1 + |x-y| + |y-z|} = \frac{|x-y|}{1 + |x-y| + |y-z|} + \frac{|y-z|}{1 + |x-y| + |y-z|} \\ &\leq \frac{|x-y|}{1 + |x-y|} + \frac{|y-z|}{1 + |y-z|}. \end{align*}