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It's posible that for any $x,y,z\in\mathbb{R}$: $$\frac{|x-y|}{1+|x-y|}\leq \frac{|x-z|}{1+|x-z|}+\frac{|z-y|}{1+|z-y|}?$$ I'm triying to prove if this inequality holds or not, but I can't find a way to start, i tried finding counter examples but the rationals won't help... Any hint?

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    Could you clarify what x, y and z are? – 2017-02-23
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    Sorry I forgot that... – 2017-02-23

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Notice that all the denominators are positive, so you can multiply out to get an equivalent inequality. After some simplification, it reduces to

$$ |x-y| \leq |x-z| + |z-y| + 2|x-z||z-y|+|x-z||z-y||y-x|$$ which is certainly true by the triangle inequality in $\mathbb{R}$, so the original inequality must also have been true.

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    Ok... It's gonna be long jaja but true, thanks a lot! – 2017-02-23
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    My wifi is bad, but in 3 minutes I'll rate your answer – 2017-02-23
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By triangle inequality, $|x-y|\le |x-z|+|z-y|$. Also the function $f(t)= \frac{t}{1+t}$ is increasing on ${\mathbb R}_+$. Therefore,

\begin{align*} \frac{|x-y|}{1+|x-y|}&=f(|x-y|)\le f(|x-z|+|y-z|)\\ & = \frac{|x-z|+|y-z|}{1+|x-z|+|y-z|}\\& \le \frac{|x-z|}{1+|x-z|} + \frac{|y-z}{1+|y-z|} \end{align*}