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Let $(X,\rho)$ be a metric space.

Is it true that $d:=\dfrac{\rho}{1+\rho}$ is also metric. If it's true can anybody give a hint how to prove the triangle inequality for $d$?

Paolo
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RFZ
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    The map $t \mapsto \frac{t}{1+t}$ is strictly increasing (on $(-1,+\infty)$). Using that together with the triangle inequality for $\rho$ helps. – Daniel Fischer Jul 06 '15 at 08:34
  • Can the triangle inequality be proven strictly? Because this problem from Rudin and the such concept like increasing function there is not in this book. – RFZ Jul 06 '15 at 08:53
  • @Pacman In mathematics, there is no difference between something that can be proven and something that can be proven strictly. If it cannot be proven strictly, it cannot be proven. Period. – 5xum Jul 06 '15 at 09:30
  • Have you tried anything at all? – anomaly Jul 09 '15 at 01:18

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Hint: Yes, $d$ is also a metric. As far as the triangle inequality is concerned, we have to prove \begin{equation*} \frac{\rho(x, y)}{1 + \rho(x, y)} \leqslant \frac{\rho(x, z)}{1 + \rho(x, z)} + \frac{\rho(z, y)}{1 + \rho(z, y)} \quad \forall \, x, y, z \in X. \end{equation*} Let $x, y, z \in X$ be taken at pleasure, and we define $A = \rho(x, y)$, $B = \rho(x, z)$ and $C = \rho(z, y)$.

So we have to prove \begin{equation*} \frac{A}{1 + A} \leqslant \frac{B}{1 + B} + \frac{C}{1 + C}, \end{equation*} which is equivalent to \begin{equation*} 1 - \frac{1}{1 + A} \leqslant 1 - \frac{1}{1 + B} + 1 - \frac{1}{1 + C}. \end{equation*} $A, B, C \geqslant 0$ and $A \leqslant B + C$, because $\rho$ is a distance function on $X$. Using these facts you can prove the inequality above.

Paolo
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