Metric triangle inequality $d_2(x,y):= \frac{d(x,y)}{d(x,y)+1}$

Question

$(X,d)$ is a metric space. $x,y,z \in X$

Now I have to proof that $(X,d_2)$ is also a metric space. To show that $d_2(x,y)=0 \leftrightarrow x=y $ and $d_2(x,y) = d_2(y,x)$ are correct was quite trivial. But with the 'triangle inequality' ($d_2(x,y) \leq d_2(x,z)+ d_2(z,y)$) I have my problems.

I have started like this:

$$ d_2(x,y) = \frac{d(x,y)}{d(x,y)+1} = 1-\frac{1}{d(x,y)+1} \leq 1-\frac{1}{1+d(x,z)+d(z,y)}$$

But I don't know how to get further. Hints are welcome :)

Hint: the function $x \mapsto \frac{x}{x+1}$ is concave-down (that is, has a negative second derivative). — Ben Grossmann, Jan 18 '15 at 23:58
Yup, you're right, but I don't see the advantage from that ^^ — Lu_kors, Jan 19 '15 at 00:01
It turns out that $f'' < 0 \implies f(a) + f(b) \geq f(a+b)$ for $a,b\ge 0$. See if you can prove this. — Ben Grossmann, Jan 19 '15 at 00:04
Yup, I can prove it, but still don't get the advantage (may due to we don't 'have' derivation till now) — Lu_kors, Jan 19 '15 at 00:30

Ben Grossmann · Accepted Answer · 2015-01-19T00:43:43.120

2

Define $f(x) = x/(x+1)$. If we can prove that $f$ is increasing and that$f(a+b) \leq f(a) + f(b)$ (which you claim to be able to do), then we can say $$ d_2(x,z) = f(d(x,z)) \le f(d(x,y) + d(y,z)) \leq \\ f(d(x,y)) + f(d(y,z)) = d_2(x,y) + d_2(y,z) $$ As desired.

edited Jan 19 '15 at 00:43

answered Jan 19 '15 at 00:37

Ben Grossmann

225,327

Too obvious ... ty – Lu_kors Jan 19 '15 at 00:42
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I just posted a duplicate, before @pizza indicated so http://math.stackexchange.com/questions/1241382/show-that-dx-y-fracx-y1x-y-is-a-metric-on-mathbbr-triangle-i/1241403#1241403 – Mirko Apr 19 '15 at 06:31

Metric triangle inequality $d_2(x,y):= \frac{d(x,y)}{d(x,y)+1}$

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