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Given $p: \mathbb{R} × \mathbb{R} \to [0,\infty)$ is defined as $$p(x,y) = \frac{|x-y|}{1+|x-y|},$$ I wish to show $p$ is a metric on $\mathbb{R}$ but I'm having difficulty proving the triangle inequality property for $p(x,y)$. So I want to show $p(x,y) \le p(x,z) + p(z,y)$. Do I make use of the standard bounded metric induced by the metric $p(x,y)=|x−y|$?

gt6989b
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Homaniac
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1 Answers1

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HINT : consider $f(t) = \frac{t}{t+1}$. What about monotonous?

openspace
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