- #1
ELESSAR TELKONT
- 44
- 0
I have the next problem:
Let [itex]d(x,y)[/itex] be a metric in [itex]\mathbb{R}^n[/itex]. If we define [itex]\tilde{d}(x,y)=\frac{d(x,y)}{1+d(x,y)}[/itex], proof that [itex]\tilde{d}(x,y)[/itex] is another metric.
I have proven that is not degenerated (i.e. [itex]\tilde{d}(x,y)=0 \longleftrightarrow x=y[/itex]) and symmetric. But I can't proof the triangle's inequality, I only get that is equivalent to proof that
[itex]\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)}[/itex]
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.
Let [itex]d(x,y)[/itex] be a metric in [itex]\mathbb{R}^n[/itex]. If we define [itex]\tilde{d}(x,y)=\frac{d(x,y)}{1+d(x,y)}[/itex], proof that [itex]\tilde{d}(x,y)[/itex] is another metric.
I have proven that is not degenerated (i.e. [itex]\tilde{d}(x,y)=0 \longleftrightarrow x=y[/itex]) and symmetric. But I can't proof the triangle's inequality, I only get that is equivalent to proof that
[itex]\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)}[/itex]
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.
Last edited: