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I am pretty sure that this is such a stupid, stupid question, but how do you prove that $d_2(x,y)={|x-y|\over {1+|x-y|}}$ satisfying the third condition to be a metric, which is the triangle inequality.

I know that I should really be able to do this, but I just can't, I keep seeing a contradiction. Please help me.

Martin Sleziak
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Akaichan
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1 Answers1

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$d_2 (x,y)+d_2 (y,z)=\frac{d(x,y)}{1+d(x,y)}+\frac{d(y,z)}{1+d(y,z)}$

$ \geq \frac{d(x,y)}{1+d(x,y)+d(y,z)}+ \frac{d(y,z)}{1+d(x,y)+d(y,z)} = \frac{d(x,y)+d(y,z)}{1+d(x,y)+d(y,z)}$

$= 1-\frac{1}{1+d(x,y)+d(y,z)} \geq 1-\frac{1}{1+d(x,z)}$

$= \frac{d(x,z)}{1+d(x,z)}=d_2 (x,z)$

M.Sina
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  • Thank you so much, it's been bothering me all night. – Akaichan Feb 08 '13 at 07:06
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    @IvordesGreenleaf : You’re welcome. – M.Sina Feb 08 '13 at 07:09
  • I just answered the same questions, after which @pizza indicated it was a duplicate. But my proof uses a bit different idea: http://math.stackexchange.com/questions/1241382/show-that-dx-y-fracx-y1x-y-is-a-metric-on-mathbbr-triangle-i – Mirko Apr 19 '15 at 06:14
  • also duplicate of http://math.stackexchange.com/questions/1109799/metric-triangle-inequality-d-2x-y-fracdx-ydx-y1?rq=1 – Mirko Apr 19 '15 at 06:27