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Proving this inequality

  1. Oct 23, 2013 #1
    1. The problem statement, all variables and given/known data
    |x + y| ≥ |x| - |y| [Hint: write out x = x + y - y, and apply Theorem 3, together with the fact that |-y| = |y|]

    2. Relevant equations
    Theorem 3: |a + b| ≤ |a| + |b|

    x = x + y - y

    |-y| = |y|

    3. The attempt at a solution

    |x + y| ≥ |x| - |y|

    x = x + y - y (don't know where to use it)

    xy ≤ |xy| = |x| |y| ( I really don't know why I am taking these steps, I am pretty much following the proof of theorem 3 in this book)

    2xy ≤ 2|x||y| ( does the |-y| = |y| come into play here to flip the inequality? )

    (x+y)^2 = x^2 + 2xy + y^2 ≤ x^2 - 2|x||y| + y^2 = (|x| - |y|)^2
     
    Last edited: Oct 23, 2013
  2. jcsd
  3. Oct 23, 2013 #2

    pasmith

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    Is it not obvious that x + y - y = x + (y - y) = x + 0 = x?
     
  4. Oct 23, 2013 #3

    tiny-tim

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    hi jimmyly! :smile:
    add some brackets …

    x = x + (y - y)

    now can you see why it's true? :wink:
    add some brackets in a different place :wink:
     
  5. Oct 23, 2013 #4

    tiny-tim

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    hi jimmyly! :smile:
    y - y = 0

    x + (y - y) = x + 0

    x + 0 = x ? o:)
     
  6. Oct 23, 2013 #5
    @mark-44: How was it false? I mean if you could clarify that would be great. Wasn't trying to spread false information or anything.
     
  7. Oct 23, 2013 #6
    I don't understand why I would write x + y ≥ x + y
    how does the subtraction turn into addition? does this mean that x + y = |x| - |y|? I see that -(-x) = + x
    but its |x| - |y| not |x| + |-y|
    or is my thinking wrong?
     
  8. Oct 23, 2013 #7
    Thank you, I understand why x = x + y - y it is very obvious! I am confused because I have no idea what it has to do with the proof and where to apply it when proving that |x+y| >= |x| - |y|
     
  9. Oct 23, 2013 #8

    Mark44

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    Tornado Dragon's post was full of errors, so I deleted it. Right at the start he tells you to remember that |x|= x. This is not true in general, particularly if x happens to be negative.



    No, it's not. You can't just ignore the absolute value signs, which Tornado Dragon seems to have done.
     
  10. Oct 23, 2013 #9

    tiny-tim

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    apply Theorem 3 to x = x + y - y (with brackets in a suitable place) :wink:
     
  11. Oct 23, 2013 #10
    @jimmyly ignore my post I have made a few mistakes in it. I will get back to you on it once I have gone it over it again.

    edit: @mark44&jimmlyl: I did make en eroneous error but I did not ignore the absolute signs like that I will redo it and post it at a later tim.e.
     
    Last edited: Oct 23, 2013
  12. Oct 23, 2013 #11
    okay so here is what i'm doing right now

    |x+y| >= |x| - |y|
    with x = x + y - y
    I got
    |x + y| >= |x| + |y| - |y| - |y|
    cancelling the |y|
    |x + y| >= |x| + |y|
    am I on the right track? :)
     
  13. Oct 23, 2013 #12
    understandable

    no worries we all make mistakes!
     
  14. Oct 23, 2013 #13

    Dick

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    Take your theorem 3, |a+b|<=|a|+|b|. Put a=x+y and b=(-y). What happens?
     
  15. Oct 23, 2013 #14

    tiny-tim

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    no!!

    Theorem 3 applies to 3 things

    use brackets to reduce the number of things to 3 !!

    then you can apply Theorem 3 :smile:
     
  16. Oct 23, 2013 #15
    so from this I got

    |a + b| <= |a| + |b|
    |(x+y) + (-y)| <= |x+y| + |-y|
    |x + y - y| <= |x+y| + |-y|
    |x| - |y| <= |x + y|
     
  17. Oct 23, 2013 #16
    oh that just did it didn't it?!
     
  18. Oct 23, 2013 #17

    Dick

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    Sure did!
     
  19. Oct 23, 2013 #18
    Would this be classified as a direct proof? I'm trying to learn proofs on my own so this is a little bit confusing to me. Thanks everyone for helping me out!
     
  20. Oct 23, 2013 #19
    Wow that's amazing. Thanks everyone! You are all wonderful
     
  21. Oct 23, 2013 #20

    tiny-tim

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    jimmyly, you seem to be worrying that there's something special about proofs

    there isn't

    if it starts with the question and finishes with the answer, it's a proof! :smile:
     
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