Prove that d(x,y)/(1+ d(x,y)) is another metric

[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.