1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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(x)-abs(y)/abs(x-y)) :a≤1
    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


    User Avatar
    Gold Member


    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


    User Avatar
    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


    User Avatar
    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted