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Help in proving abs(abs(x)-abs(y))<=abs(x-y)

  1. Apr 2, 2014 #1
    1. The problem statement, all variables and given/known data
    already proved abs(x)-abs(y)≤abs(x-y)
    I just don't know how to prove it
    2. Relevant equations

    3. The attempt at a solution
    we have :abs(x)-abs(y)≤abs(x-y)
    abs(x)-abs(y)/abs(x-y)≤1 / abs(x-y) is diff than 0

    proving abs(abs(x)-abs(y))/abs(x-y))≤1 :abs(x-y)diff than 0 then...
    abs(abs(x)-abs(y)/abs(x-y))≤1 / abs(x-y)=abs(abs(x-y))

    a<=abs(a)
    a=abs(x)-abs(y)/abs(x-y)) :a≤1
    abs(x)-abs(y)/abs(x-y))≤abs(abs(x)-abs(y)/abs(x-y))....
    can't continue from here...I don't think this is the right way to do it
    please any help.
     
  2. jcsd
  3. Apr 2, 2014 #2

    MathematicalPhysicist

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    Gold Member

    Well,

    from |x| <= |x-y|+|y|
    and |y| <= |y-x|+|x|

    So you have: |x|-|y| , |y|-|x| <= |x-y|
    Then because ||x|-|y|| equals one of them, what can you infer?
     
  4. Apr 2, 2014 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    I would do this in four cases:
    1) x> 0, y> 0
    2) x> 0, y< 0
    3) x< 0, y> 0
    4) x< 0, y< 0

    And each case would have two "subcases"
    a) |x|> |y|
    b) |x|< |y|
     
  5. Apr 3, 2014 #4

    Curious3141

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    Homework Helper

    The way I did it was to take ##\displaystyle ||x| - |y||.||x| + |y||## and simplify it algebraically, ending up with ##\displaystyle ||x| - |y|| = |x-y|.|\frac{x+y}{|x| + |y|}|##.

    It should be quite trivial to observe that the factor on the right is always positive, and has a maximum value of 1, and that only occurs when x and y are of the same sign.
     
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