How can I prove this is a metric?

My inequality chain was:

\begin{align*} d(x,z)+d(z,y)={}&\sum_{n=1}^\infty2^{-n}\frac{\rho_n(x-z)}{1+\rho_n(x-z)}+\sum_{n=1}^\infty2^{-n}\frac{\rho_n(z-y)}{1+\rho_n(z-y)}={} \\ {}={}&\sum_{n=1}^\infty2^{-n}\left(\frac{\rho_n(x-z)}{1+\rho_n(x-z)}+\frac{\rho_n(z-y)}{\rho_n(z-y)+1}\right)={} \\ {}={}&\sum_{n=1}^\infty2^{-n}\frac{\rho_n(x-z)+2\rho_n(x-z)\rho_n(z-y)+\rho_n(z-y)}{1+\rho_n(x-z)+\rho_n(z-y)+\rho_n(x-z)\rho_n(z-y)}\geq{} \\ {}\geq{}&\sum_{n=1}^\infty2^{-n}\frac{\rho_n(x-y)}{1+\rho_n(x-z)+\rho_n(z-y)+\rho_n(x-z)\rho_n(z-y)}, \end{align*}

where the first = is by definition, the second one is because the series converge (they are, in general, less than or equal to the geometric series of ratio $\frac12$ which converges) so I can permute the terms, the third one simply sums the fractions, and the inequality throws away some positive terms from the numerator. If I only remove the two, I have a sum of $2^{-n}\frac{\alpha_n}{1+\alpha_n}$, where:

$$\alpha_n=\rho_n(x-z)+\rho_n(z-y)+\rho_n(x-z)\rho_n(z-y)\geq\rho_n(x-y)$$

by the triangle inequality on the first two terms and throwing away the third which is positive, so those fractions are, since as Daniel pointed out $\frac{t}{1+t}$ is strictly increases, strictly greater than $\frac{\rho_n(x-y)}{1+\rho_n(x-y)}$, which completes the proof.

Thomas's comment notes something that seems not too useful here, but is anyway interesting: if $d$ is a metric, $d'(x,y)=\frac{d(x,y)}{1+d(x,y)}$ is too. That is because the triangle inequality is proved as below, and the other properties of a metric are obvious.

$$d'(x,z)+d'(z,y)=\frac{d(x,z)}{1+d(x,z)}+\frac{d(z,y)}{1+d(z,y)}=\frac{d(x,z)+d(z,y)+2d(z,x)d(z,y)}{1+d(x,z)+d(z,y)+d(x,z)d(z,y)},$$

which as before is at least $\frac{d(x,y)}{1+d(x,y)}=d'(x,y)$, since $\frac{t}{1+t}$ is strictly increasing.

And $\frac{t}{1+t}$ is strictly increasing on all of $(-1,+\infty)$ because its derivative is:

$$\frac{1}{1+t}-\frac{t}{(1+t)^2}=\frac{1}{(1+t)^2}>0,$$

for all $t>-1$.