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Prove of new metric space

  1. Mar 18, 2012 #1
    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??
     
  2. jcsd
  3. Mar 18, 2012 #2
    Can you prove this assuming a is bigger than both b and c? Can you prove it assuming a is not the biggest?
     
  4. Mar 18, 2012 #3
    What do you mean? I know a≤b+c...

    So I assume a is bigger than both a and c?
     
  5. Mar 19, 2012 #4
    I mean that if you can prove an assertion under some condition, and then under the complementary condition, then you are done. And I have also told you what condition works well in your case.
     
  6. Mar 19, 2012 #5
    This is natural to assume given the triangle inequality for the original metric.
     
  7. Mar 19, 2012 #6
    Let:
    a=d(x,y)
    b=d(x,z)
    c=d(z,y)

    a[itex]\leq[/itex]b+c
    This is the triangle inequality for the original metric right?

    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] -- (*)

    Does this means that I try to prove (*) by two cases:
    1. assuming a>b and a>c
    2. assuming a<b and a<c???
     
  8. Mar 19, 2012 #7
    I'm not sure whether this way of proving it is okay:

    I take a look at the function

    f(x)=[itex]\frac{x}{\sqrt{1+x}}[/itex]

    f'(x)=[itex]\frac{2+x}{2(1+x)^{\frac{3}{2}}}[/itex]
    f'(x)>0 if and only if x≥-1.

    Since I know that 0≤a≤b+c, f(x) is an increasing function in the interval [0,b+c]. Therefore, by choosing a=b+c, f(a)=[itex]\frac{a}{\sqrt{1+a}}[/itex]=[itex]\frac{b+c}{\sqrt{1+b+c}}[/itex] is the maximum possible value.

    Hence it is enough to show that [itex]\frac{b+c}{\sqrt{1+b+c}}[/itex][itex]\leq[/itex][itex]\frac{b}{\sqrt{1+b}}[/itex]+[itex]\frac{c}{\sqrt{1+c}}[/itex]

    Since 1+b+c≥1+b and 1+b+c≥1+c,
    [itex]\frac{b+c}{\sqrt{1+b+c}}[/itex]=[itex]\frac{b}{\sqrt{1+b+c}}[/itex]+[itex]\frac{c}{\sqrt{1+b+c}}[/itex][itex]\leq[/itex][itex]\frac{b}{\sqrt{1+b}}[/itex]+[itex]\frac{c}{\sqrt{1+c}}[/itex]

    Therefore,
    [itex]\frac{a}{\sqrt{1+a}}[/itex][itex]\leq[/itex][itex]\frac{b}{\sqrt{1+b}}[/itex]+[itex]\frac{c}{\sqrt{1+c}}[/itex]

    It seems a bit dodgy :bugeye:
     
  9. Mar 19, 2012 #8
    Your solution is good. You use both the idea about f'>0 and the analysis of the denomintarors in one argument.
     
  10. Mar 19, 2012 #9
    So is this method valid?
     
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