How to show Triangle inequality

Question

Let $X$ be a normed linear space and a function $p$ on $X$ given by $p(x)=\frac {||x||} {1+||x||}$. Let a metric $r$ on $X$ be given by $r(x,y)= p(x-y)$.Prove that it is a metric on $X$. I understand except for triangle inequality, but I don’t know how to show $r$ satisfies triangle inequality.

Answer 1

Hint: On $[0,\infty)$ the function $\frac x {1+x}$ is an increasing and a simple algebraic argument shows that $\frac {x+y} {1+x+y} \leq \frac x {1+x}+\frac y {1+y}$. This gives triangle inequality.

Details for triangle inequality: let $f(x)=\frac x {1+x}, A=\|x-y\|, B=\|y-z\|$ and $C=\|z-y\|$. Then $A\leq B+C$. Since $f$ is increasing we get $f(A) \leq f(B+C)$. Since $f(x+y) \leq f(x)+f(y)$ we get $f(A) \leq f(B)+f(C)$.