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Homework Help: Triangle Inequality Proof help

  1. Sep 21, 2010 #1
    1. The problem statement, all variables and given/known data

    1] l x + y l < or equal to l x l + l y l




    2. Relevant equations

    x^2 + 2xy + y^2


    3. The attempt at a solution

    Left side. i Squared left side to begin with, and i got x^2 + 2xy + y^2 and also did the same for the right side, but it would have absolute sign between each variable.

    x^2 + 2xy + y^2 < or equal to lxl^2 + 2lxllyl + lyl^2

    so its just (x + y)^2 < or equal to ( lxl+lyl )^2
    not sure what im supppose to do next..

    it just seems like im doing somethign and undoing something without any purpose.
    can anyone clear this from my head?
     
  2. jcsd
  3. Sep 21, 2010 #2

    Dick

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    The point is that 2xy<=2|x||y| and x^2=|x|^2 and y^2=|y|^2.
     
  4. Sep 21, 2010 #3
    so if it asked for

    lA - Bl <\ lAl + lBl

    We would get the same as above but except it would be Sqrt[(a - b)^2] <\ lAl + lBl

    or is it Sqrt[(a - b)^2] = lAl + lBl
    as a final answer?
     
  5. Sep 21, 2010 #4

    Dick

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    If you are supposed to prove |A-B|<=|A|+|B|, the final answer would be an argument proving it. Can you give a proof?
     
  6. Sep 21, 2010 #5
    lA-Bl^2 <= ( lAl +lBl ) ^2

    A^2 - 2AB + B^2 <= lAl^2 + 2lAllBl + lBl^2

    -2AB <= 2lAllBl

    would this be correct?
    im not sure if i can subtract absolute value to a non absolute value...
    still confused.
     
  7. Sep 21, 2010 #6

    Dick

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    -2AB is either equal to 2|A||B| or -2|A||B| depending on the signs of A and B, right? In either case, -2AB is less than or equal to 2|A||B|. Is that the part you are asking about?
     
  8. Sep 21, 2010 #7
    yea doesnt that prove |A-B|<=|A|+|B|

    square both sides.
    lA-Bl^2 <= ( lAl +lBl ) ^2

    A^2 - 2AB + B^2 <= lAl^2 + 2lAllBl + lBl^2

    you subtract A^2 and B^2 from both sides.
    your left with -2AB <= 2lAllBl which is true.

    or is this not the final stage?
     
  9. Sep 21, 2010 #8

    Dick

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    Sure. Except to really present it as a proof you probably want to start with -2AB<=2|A||B|. Now add A^2 and B^2 to both sides, factor, note both sides are nonnegative and take the square root. In other words do the derivation backwards. But you certainly have the right idea.
     
  10. Sep 21, 2010 #9
    what do you mean by adding a^2 and b^2 to both sides, you mean remove the absolute sign from A and B from the right side?

    so my solution above was not good enough as a proof right ?
    I need to make sure that my answer is at a reasonable level to proove the inequality.

    so now i add A^2 and B^2.
    A^2 - 2AB + B^2 <= A^2 + 2lAllBl + B^2

    and take the square root ? how?
     
  11. Sep 21, 2010 #10

    Dick

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    A^2=|A|^2. Ditto for B. Now FACTOR both side into squares. Then take the square root.
     
  12. Sep 21, 2010 #11
    A^2 - 2AB + B^2 <= A^2 + 2lAllBl + B^2
    (A-B)^2 <= (lAl+lBl)^2
    (A-B) <= lAl + lBl

    uh, now we're back to the original inequality, is this what we're suppose to do ?
    do somethign and undo it again ? im new to prooving inequalities ,
    is this correct?
     
  13. Sep 21, 2010 #12

    Dick

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    It's not a question of doing and undoing something. You 'discovered' the proof by squaring both sides and reducing it to -2AB<=2|A||B|. Which you know is true. So now to prove it you start with what you know is true. Your first line should be -2AB<=2|A||B|. 'Proofs' usually start with what you know, then end up with what you want to prove.
     
  14. Sep 21, 2010 #13
    -2AB <= 2lAllBl
    A^2 - 2AB + B^2 <= A^2 + 2lAllBl + B^2
    (A - B)^2 <= (lAl + lBl)^2
    (A - B) <= lAl + lBl

    So it would look something like this ?
     
  15. Sep 21, 2010 #14

    Dick

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    It would look exactly like that. If you want to be fancy you could add a comment to each line justifying the step you took. But I think that's good enough.
     
  16. Sep 21, 2010 #15
    awesome, thanks for the awesome help !
    i just didnt grasp the concept of prooving inequality at first.
     
  17. Sep 24, 2010 #16
    lets say this question was in an our midterm exam.
    how would i go about justifying ?

    Heres my shot at it.

    Since we know that 2lAlBl >= 0 for any real number. Therefore, it is greater than
    -2AB which yields a negative value for the following 3 cases.

    A < 0 , B > 0
    A > 0, B < 0
    A >0, B >0

    By adding A^2 and B^2 to both sides, we can factor and squareroot the equation.
    which gives us (A - B) <= lAl + lBl

    (One question, is it resonable to just start with -2AB <= 2lAllBl ? )
    curious..
     
  18. Sep 24, 2010 #17

    Dick

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    -2AB<=2|A||B| in ALL cases. Not just the three you listed. It's 'greater than or equal to' not just 'greater than'.
     
  19. Sep 24, 2010 #18
    ok so what is it that im justifying at the end or in the middle.

    either 2lalbl >= -2ab

    or that a - b <= lal + lbl

    we start with 2lalbl >= -2ab and end with a - b <= lal + lbl, which is what its asking us to proove.

    please help me understand the world of inequality.
     
  20. Sep 24, 2010 #19

    Dick

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    That might depend on what you are trying to prove. I'm getting all confused. You started out trying to prove |a+b|<=|a|+|b|, right? Did the problem change? Why don't you state what you are trying to prove and the steps you might use to get there and I'll see if I can help.
     
    Last edited: Sep 24, 2010
  21. Sep 24, 2010 #20
    yea i started out trying to prove la+bl <= lal + lbl

    this is exactly how i solved it.

    "
    (a-b)^2 <= (lal + lbl)^2
    a^2 - 2ab + b^2 <= lal^2 + 2lalbl + lbl^2
    a^2 - 2ab + b^2 <= a^2 + 2labl + b^2
    Examine the value of -2ab and 2labl. 2labl >= -2ab is true for all real
    values since labl >= ab

    Take the square root of a^2 - 2ab + b^2 <= a^2 + 2labl + b^2
    we have l a - b l <= lal + lbl
    "

    That is what I had in mine. I wrote it as if this question was on the exam or an assignment. Please i need to know where i went wrong, starting from the beggining of quotation, please let me know whre to fix my errors.
     
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