Let be $\rho (x,y) $ a metric on X. Prove that $\rho _1 (x,y)=\rho (x,y)/(1+\rho (x,y) ) $ is a metric on X.

Question

I'm trying to do this problem. Let be $\rho (x,y) $ a metric on X. Prove that $$\rho _1 (x,y)=\frac{\rho (x,y)}{1+\rho (x,y)}$$ is a metric on $X$.
I have problems to prove the triangle inequality because I have this: I know that $\rho (x,y) $ is a metric $\Rightarrow $ $\rho (x,z)\leq \rho (x,y)+\rho (y,z)$ then $ \frac{1}{\rho (x,z) }\geq \frac{1}{\rho (x,y)+\rho (y,z)}$ and the inequality changes, I need to conserve it, can someone help me please?.

Answer 1

Firstly, from $\rho(x,z)\le \rho (x,y)+\rho(y,z)$ you can conclude that $\frac{1}{\rho(x,z)}\ge \frac{1}{\rho (x,y)+\rho(y,z)}$, but absolutely not the inequality you wrote. A hint to solve the exercise is to think of it more generally. Seek conditions on a function $f\colon \mathbb R_+ \to \mathbb R_+$ that will guarantee that if $\rho$ is a metric on some set, then $\tau(x,y)=f(\rho(x,y))$ is also a metric. Your case is $f(t)=\frac{t}{1+t}$. Hint: concavity!