Let $(E, D)$ be a metric space.
Consider $D_1: E\times E \to \mathbb{R}$ where $$ D_1(x,y)=\frac{D(x,y)}{1+ D(x,y)}. $$
I read some note about it but I want to find why $D_1$ is also a metric and $$ D_1(x,y) \le \min \{ 1, D(x, y)\}. $$
Could anyone help me?
Two of the three properties of a metric are obviously satisfied by $D_1$, namely that
because they hold for $D$. Also, from $$ \frac{a}{1+a}\le 1,\qquad \frac{a}{1+a}\le a $$ which are easily verified for $a\ge0$, you can derive $$ \frac{a}{1+a}\le \min\{1,a\} $$ Now set $a=D(x,y)$ and you have the last inequality.
The most difficult property is the triangle inequality, that is $$ D_1(x,z)\le D_1(x,y)+D_1(y,z) $$ or $$ \frac{c}{1+c}\le\frac{a}{1+a}+\frac{b}{1+b} $$ where $c=D(x,z)$, $a=D(x,y)$ and $b=D(y,z)$, where we know that $c\le a+b$ by the triangle inequality satisfied by $D$.
We can clear denominators, because we know that $a\ge0$, $b\ge0$ and $c\ge0$, so we have to prove that $$ c(1+a)(1+b)\le a(1+b)(1+c)+b(1+a)(1+c) $$ or $$ c+ac+bc+abc\le a+ab+ac+abc+b+ab+bc+abc $$ that becomes $$ c\le a+2ab+b+abc $$ that will be verified as soon as we prove the stronger inequality $$ a+b\le a+2ab+b+abc $$ that's of course true.