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This is related to Proof that every metric space is homeomorphic to a bounded metric space but I can remember that if $d$ is a metric, then $d'(x,y):=\frac{d(x,y)}{1+d(x,y)}$ is also a metric that defines the same topology.

I'm stuck on how to prove that $d'$ satisfies the triangle in-equality Am I just missing the trick how to prove it, or does my memory play tricks on me?

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Gyro Gearloose
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    Some related older posts: http://math.stackexchange.com/questions/309198/if-d-is-a-metric-then-d-1d-is-also-a-metric http://math.stackexchange.com/questions/686792/showing-rho-x-y-fracdx-y1dx-y-is-a-metric-space http://math.stackexchange.com/questions/310612/show-that-d-bx-y-fracdx-y1dx-y-is-a-metric – Martin Sleziak Jan 27 '16 at 20:25

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Define $\phi:[0,\infty)\to[0,\infty)$ by $$\phi(t)=\frac{t}{1+t}.$$Since $\phi$ is increasing it's enough to show that $$\phi(a+b)\le\phi(a)+\phi(b)\quad(a,b\ge0).$$This is the same as $$\phi(a+b)-\phi(a)\le\phi(b)-\phi(0),$$which is the same as $$\int_0^b(\phi'(t+a)-\phi'(t))\,dt\le0.$$So it's enough to show that $\phi'$ is decreasing.

David C. Ullrich
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