Show that $d^\ast$ is a metric For $x$ and $y$ in $R$, let $d(x,y)$ be a metric. Show that $$d^\ast(x,y)=\frac{d(x,y)}{1+d(x,y)}$$ is also a metric.
It is fairly straightforward to show that 


*

*$d^\ast(x,y)=0$ if $x=y$

*$d^\ast(x,y)=d^\ast(y,x)$ for every $x,y \in R$  


However, I was stuck while trying to show that


*

*$d^\ast(x,y)\le d^\ast(x,z)+d^\ast(z,y)$ for every $x,y,z \in R$


I tried to simplify: $$1-\frac{1}{1+d(x,z)}\le2-\frac{1}{1+d(x,y)}-\frac{1}{1+d(y,z)}$$ but then fail to advance...
 A: To simplify notations let $d(x,y)=a,d(y,z)=b,d(z,x)=c$.So we know that $a\le b+c$. Now
$${b\over1+b}+{c\over1+c}-{a\over 1+a}={b(1+c+a+ca)+c(1+a+b+ab)-a(1+b+c+bc)\over(1+a)(1+b)(1+c)}={b+c-a+\text{non-negative terms}\over(1+a)(1+b)(1+c)}\ge 0$$
A: Let $d(x, y) = a, d(y, z) = b, d(x, z) = c$.
First suppose that $c \leq a$ or $c \leq b$. We can assume the first one without loss of generality. Then
$$
c(1 + a) \leq a(1 + c) \implies \frac{c}{1 + c} \leq \frac{a}{1 + a}\\
\implies d^*(x, z) \leq d^*(x, y) \implies d^*(x, z) \leq d^*(x, y) + d^*(y, z).
$$
Now suppose both $c \geq a$ and $c \geq b$. Without loss of generality we can assume $a \leq b \leq c$. Then $a \leq b \implies \frac{1}{1 + b} \leq \frac{1}{1 + a}$. Also from the triangle inequality of the metric $d$, we have $c \leq a + b \implies 0 \leq c - b \leq a$. Multiplying both inequalities gives
$$
\frac{c - b}{1 + b} \leq \frac{a}{1 + a} \implies \frac{c}{1 + c} \leq \frac{c}{1 + b} \leq \frac{a}{1 + a} + \frac{b}{1 + b}
$$
since $b \leq c$. Hence $d^*(x, z) \leq d^*(x, y) + d^*(y, z)$.
