Proving the triangle inequality for the metric d(x,y)/(1+d(x,y))

[ELESSAR TELKONT]

I have the next problem:

Let $d(x,y)$ be a metric in $\mathbb{R}^n$. If we define $\tilde{d}(x,y)=\frac{d(x,y)}{1+d(x,y)}$, proof that $\tilde{d}(x,y)$ is another metric.

I have proven that is not degenerated (i.e. $\tilde{d}(x,y)=0 \longleftrightarrow x=y$) and symmetric. But I can't proof the triangle's inequality, I only get that is equivalent to proof that

$\frac{d(x,y)}{1+d(x,y)}\leq \frac{d(x,z)}{1+d(x,z)}+\frac{d(y,z)}{1+d(y,z)}$

I need some help, please, because if the denominators wouldn't exist it will be easy. That's I need to know if the denominators in the right member of inequality are lesser than the one on the other member. I hope can you help me.

 

[robphy]

Did you try to multiply through to clear all of the denominators? [multiply the inequality by the product of all of the denominators]
It seems that there will be numerous terms that can be grouped.
Make use of the fact that d already satisfies the inequality.

 

[quasar987]

This is called the bounded metric (associated with d) because it is bounded by 1.

I did this problem earlier this summer, and I found it helpful to prove this lemma first:

"If $0\leq a \leq b$, then $$\frac{a}{1+a}\leq \frac{b}{1+b}$$"

Then knowing that d already satisfies the triangle inequality, the answer is very near.

 

[ZioX]

Quasar is essentially is saying that $f(x)=\frac{x}{1+x}$ is increasing on $\mathbf{R}^+$. You can prove this algebraically, which I believe quasar is hinting at, but it "easier" to do it with calculus.