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Math Help - fiddling with inequalities (quick question)

  1. #1
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    fiddling with inequalities (quick question)

    If I start with the triangle inequality:

    ||x-z|+|y-z|| \ge |x-y|

    and then I switch the sign on the LHS, which is true:


    ||x-z|-|y-z|| \le |x-y|

    or

    ||x-z|-|y-z|| = |x-y|
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  2. #2
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    I suppose obviously the second one...

    So how would I prove:

    ||x-z|-|y-z|| \le |x-y| for all x, y, z \in X ?
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  3. #3
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    Start with:

    x-z=t

    y-z=s

    x-y=t-s
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  4. #4
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    Quote Originally Posted by MichaelMath View Post
    If I start with the triangle inequality:
    ||x-z|+|y-z|| \ge |x-y|
    Start with |x|\le |x-y|+|y| implies |x|-|y|\le |x-y|

    Likewise |y|-|x|\le |y-x|=|x-y|

    So -|x-y|\le |x|-|y|\le |x-y|.

    That implies \left||x|-|y|\right|\le |x-y|
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  5. #5
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    Thanks mate, I figured it out using Zarathustra's suggestion. Can you get me started on this one?

    |d(x,z)-d(y,u)| \leq d(x,y)+d(z,u)
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  6. #6
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    Start by showing d(x,z)-d(z,y)\le d(x,y)~\&~d(y,z)-d(y,u)\le d(z,u).

    Note that d(y,z)=d(z,y) and add those to together.
    Last edited by Plato; October 20th 2010 at 01:26 PM.
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  7. #7
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    I need to read a bit about manipulating inequalities and absolute values... any suggestions?
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  8. #8
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    Quote Originally Posted by MichaelMath View Post
    I need to read a bit about manipulating inequalities and absolute values... any suggestions?
    Can you get
    d(x,z)-d(y,u)\le d(x,y)+d(u,z)~\&~d(y,u)-d(x,z)\le d(u,z)+d(x,y)?

    Do you understand that if a>0~\&~-a<b<a then |b|< a?
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  9. #9
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    Okay...

    d(x,z)-d(z,y)\le d(x,y)

    d(y,z)-d(y,u)\le d(z,u)

    So:

    <br />
d(x,z)-d(z,y)+d(y,z)-d(y,u) \leq d(x,y)+d(z,u)

    d(x,z)-d(y,u) \leq d(x,y)+d(z,u)

    and the second one...
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  10. #10
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    ok, i got it.
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