Suppose that (X,d) is a metric(adsbygoogle = window.adsbygoogle || []).push({});

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??

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# Prove of new metric space

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