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This is a follow up to a well known question

Showing $\rho (x,y)=\frac{d(x,y)}{1+d(x,y)}$ is a metric

A general proof is as follows: Let $x,y,z \in (X, \rho)$

\begin{align} \rho(x,z) &= \dfrac{d(x,z)}{1+d(x,z)}\\ & \leq \dfrac{d(x,y)+d(y,z)}{1+d(x,y)+d(y,z)}\\ &= \dfrac{d(x,y)}{1+d(x,y)+d(y,z)} + \dfrac{d(y,z)}{1+d(x,y)+d(y,z)} \\&\leq \rho(x,y) + \rho(y,z) \end{align}

This might be very simple but how do you rigorously argue that

$$\dfrac{d(x,z)}{1+d(x,z)}\leq \dfrac{d(x,y)+d(y,z)}{1+d(x,y)+d(y,z)}$$

If only the numerator guy got bigger due to triangle inequality, then I would have no problem. However, the thing in the denominator also got bigger. How do we know that this inequality holds?

Martin Sleziak
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Fraïssé
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  • Cross multiply then use the triangle inequality –  Jul 15 '16 at 20:43
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    I'd say $f(x) = \frac{x}{1+x}$ is concave on $[0,\infty)$ : $f(a) \le f(b) + f(c)$ whenever $a \le b+c$ – reuns Jul 15 '16 at 20:50
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    Let $f(x)=\frac{x}{x+1}$. We have $f'(x)=\frac{1}{(x+1)^2}>0$, so f is strictly increasing and preserves inequality. Therefore $d(x,z)\leq d(x,y)+d(y,z)$ implies inequality $f(d(x,z))\leq f(d(x,y)+d(y,z))$ – alans Jul 15 '16 at 21:01

2 Answers2

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$0 \le a < b \implies 0 \le \frac{a}{b} \le\frac{a+n}{b+n}< 1$ where $n\ge0$ with $\lim_{n\to\infty}\frac{a+n}{b+n}=1$

We can use this since, of course, the difference between the numerators is the same as the difference between the denominators.

see this question if you want a proof of the inequality.

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n1000
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One general reason, as user1952009 notes in the comments, is that $f(x) = \frac{x}{1 + x}$ is increasing and concave on $[0, \infty)$.

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Andrew D. Hwang
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