Show that $d'$ is a metric

Question

Let $(X,d)$ a metric space. Show that $d'(x,y)=\dfrac{d(x,y)}{1+d(x,y)}$ define a metric on $X$.

Hello, I have this problem. The triangular inequality is my difficulty. Indeed, we have that show that $d'(x,z)\le d'(x,y)+d'(y,z),\forall x,y,z\in X.$ Good, I do this...

$d'(x,z)=\dfrac{d(x,z)}{1+d(x,z)}\le \dfrac{d(x,y)+d(y,z)}{1+d(x,z)}=\dfrac{d(x,y)}{1+d(x,z)}+\dfrac{d(y,z)}{1+d(x,z)}.$

I don't understand how does with $d(x,z)$? Or if exists any form for bounded this expression.

For clarity, we want to show that for $a,b,c\geqslant$, $a+b \leqslant c$ gives us $\frac{a}{1+a}+\frac{b}{1+b}\leqslant \frac{c}{1+c}$. — daruma, May 02 '17 at 09:30
the inequality is the other way around in the previous comment — JJR, May 02 '17 at 09:55

score 3 · Answer 1 · answered May 02 '17 at 09:28

I found it easier to start from the other end:

$\begin{align*} d'(x,y)+d'(y,z) & = \frac{d(x,y)}{1+d(x,y)}+\frac{d(y,z)}{1+d(y,z)}\\ & = \frac{d(x,y)+d(y,z)+2d(x,y)d(y,z)}{1+d(x,y)+d(y,z)+d(x,y)d(y,z)}\\ & \geq \frac{d(x,y)+d(y,z)}{1+d(x,y)+d(y,z)}\\ & \geq \frac{d(x,z)}{1+d(x,z)}. \end{align*}$

The first inequality is since $d(x,y)$ and $d(y,z)$ are both nonnegative and

$\frac{p+2c}{1+p+c}\geq \frac{p}{1+p} \Longleftrightarrow 2+2p\geq p$,

which is clearly true for nonnegative constants $p$ and $c$.

Do you think you can get the last inequality?

Show that $d'$ is a metric

1 Answers1