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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 $\displaystyle \frac{(1+x+z+xz+y+y^2)}{(1+y)}\ge 1+x+z$ ?

Please help me in proving this , otherwise any easiest way for triangle inequality ?

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    $\begingroup$ Is the proposition correct? I’ve got that $d(x,y)+d(y,z)<d(x,z)$ when $x>y>z$. $\endgroup$
    – Jihai Zhu
    Mar 7 at 7:37
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    $\begingroup$ 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 $[0,+\infty)$ $\endgroup$ Mar 7 at 11:35
  • $\begingroup$ 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 $z$?? $\endgroup$
    – runway44
    Mar 7 at 11:45
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It's not a metric. $d(1,3)>d(1,2)+d(2,3)$.

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