Show $\rho (x,y)=\dfrac{d(x,y)}{1+d(x,y)}$ is a metric on the metric space $X$, equipped with the Euclidean metric $d$.
I've already shown that the positivity $\rho(x,y)\geq 0$, the symmetry $\rho(x,x)=\rho(y,y)=0$.
I'm having trouble proving the last condition of a metric space though: the triangle inequality must be valid:
$\rho(x,y)\leq \rho(x,p)+\rho (p,y)$, for $p\in X$. So, $\dfrac{d(x,y)}{1+d(x,y)}\leq \dfrac{d(p,x)}{1+d(p,x)}+\dfrac{d(p,y)}{1+d(p,y)}$
Could someone provide a hint as to get started? No complete solutions please. Thanks in advance.
Best Answer
First show that $\frac{a}{1+a}\le \frac{b}{1+b},$ if $0\le a\le b$. Secondly, try to prove that $\frac{a+b}{1+c+d}\le\frac{a}{1+c}+\frac{b}{1+d}$, if $0\le a,b,c,d$. And thirdly, use that your function $d(x,y)$ is a metric.