- #1
Lily@pie
- 109
- 0
Suppose that (X,d) is a metric
Show [itex]\tilde{d}[/itex](x,y) = [itex]\frac{d(x,y)}{\sqrt{1+d(x,y)}}[/itex] is also a metric
I've proven the positivity and symmetry of it.
Left to prove something like this
Given a[itex]\leq[/itex]b+c
Show [itex]\frac{a}{\sqrt{1+a}}[/itex][itex]\leq[/itex][itex]\frac{b}{\sqrt{1+b}}[/itex]+[itex]\frac{c}{\sqrt{1+c}}[/itex]
I try to prove this
a[itex]\sqrt{1+b}[/itex][itex]\sqrt{1+c}[/itex]=b[itex]\sqrt{1+a}[/itex][itex]\sqrt{1+c}[/itex]+c[itex]\sqrt{1+a}[/itex][itex]\sqrt{1+b}[/itex]
but I'm just stuck!
cz previously I've proven this [itex]\frac{a}{1+a}[/itex][itex]\leq[/itex][itex]\frac{b}{1+b}[/itex]+[itex]\frac{c}{1+c}[/itex] before...
I guess I can't use the same method??
Show [itex]\tilde{d}[/itex](x,y) = [itex]\frac{d(x,y)}{\sqrt{1+d(x,y)}}[/itex] is also a metric
I've proven the positivity and symmetry of it.
Left to prove something like this
Given a[itex]\leq[/itex]b+c
Show [itex]\frac{a}{\sqrt{1+a}}[/itex][itex]\leq[/itex][itex]\frac{b}{\sqrt{1+b}}[/itex]+[itex]\frac{c}{\sqrt{1+c}}[/itex]
I try to prove this
a[itex]\sqrt{1+b}[/itex][itex]\sqrt{1+c}[/itex]=b[itex]\sqrt{1+a}[/itex][itex]\sqrt{1+c}[/itex]+c[itex]\sqrt{1+a}[/itex][itex]\sqrt{1+b}[/itex]
but I'm just stuck!
cz previously I've proven this [itex]\frac{a}{1+a}[/itex][itex]\leq[/itex][itex]\frac{b}{1+b}[/itex]+[itex]\frac{c}{1+c}[/itex] before...
I guess I can't use the same method??