prove that $d(x,y):=\left|\frac x{1+y}-\frac y{1+x}\right|$ forms a metric on $(0,\infty)$

Let $$X=(0,\infty)$$. Define, $$\displaystyle d(x,y):=\left|\frac x{1+y}-\frac y{1+x}\right|$$. Show that $$d$$ forms a metric on $$X$$.

I have done all the conditions except triangle inequality. I've stuck here! Can anyone give some hint for triangle inequality ?

We have,

\begin{align*} d(x,z)&=\left|\frac x{1+y}-\frac y{1+x}\right|\\ &=\left|\frac{(x-z)(1+x+z)}{(1+x)(1+z)}\right|\\ &\le \left|\frac{(x-z)(1+x+z+xz+y+y^2)}{(1+x)(1+y)(1+z)}\right| \tag 1 \end{align*} If we can write the inequality (1) then I can prove the result. But how to prove that $$\displaystyle \frac{(1+x+z+xz+y+y^2)}{(1+y)}\ge 1+x+z$$ ?

• Is the proposition correct? I’ve got that $d(x,y)+d(y,z)<d(x,z)$ when $x>y>z$. Mar 7 at 7:37
• I would have expected $d(x,y):= \left|\frac{x}{1+x} - \frac{y}{1+y}\right|$ instead. This is a well-known metric on $[0,+\infty)$ Mar 7 at 11:35
• In (1), how did you start with $d(x,z)$, which involves no $y$, to get a formula that involves $y$ but no $z$, followed by a formula that involves $z$ and no $y$ again, and end at a formula that has both $y$ and $z$?? Mar 7 at 11:45
It's not a metric. $$d(1,3)>d(1,2)+d(2,3)$$.