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Suppose $f,g,h>0$ for every $x$. How can I show this inequality $$ \frac{|f(x)-g(x)|}{1+|f(x)-g(x)|}\le \frac{|f(x)-h(x)|}{1+|f(x)-h(x)|}+\frac{|h(x)-g(x)|}{1+|h(x)-g(x)|} $$ Adding subtracting $h(x)$ to numerator is the way to go for numerator part but what about denominator?

FreeMind
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    Are you sure in the RHS of your inequality is a multiplication and not an addition ? (I strongly susêct that you want to prove that $d(x,y)=\frac{|x-y|}{1+|x-y|}$ is a metric). – Surb Mar 09 '21 at 12:43
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    ... and that has been asked and answered many times before, e.g. here: https://math.stackexchange.com/q/686792. – Martin R Mar 09 '21 at 12:47
  • @Surb My bad man. Edited the question. Yeah I exactly want to prove that $d$ is metric – FreeMind Mar 09 '21 at 13:20
  • @MartinR The accepted answer over there is just stating the problem. I know what I have to show but how! – FreeMind Mar 09 '21 at 14:33
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    @FreeMind: Here is another one: https://math.stackexchange.com/q/981684 – you can find many more with Approach0 – Martin R Mar 09 '21 at 14:45

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