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Homework Help: How to justify?

  1. Sep 24, 2010 #1
    1. The problem statement, all variables and given/known data

    la - bl <= lal + lbl

    2. Relevant equations

    3. The attempt at a solution

    I have prooved this triangle inequality. but please check if there are any obvious errors.

    -2ab <= 2labl
    a^2 - 2ab + b^2 <= a^2 + 2labl + b^2
    ( la - bl)^2 <= (lal + lbl)^2

    *sqrt both sides
    la - bl <= lal + lbl

    The problem is now that i have somewhat prooved this inequality, how do i justify it?
    In an assignment or on a test I need to write the process of my work.
    How would i go aobut justifying this proof?

    Heres my shot at it:

    ab <= l ab l is true for all real numbers
    therefore -2ab <= 2l ab l is the same case.

    add a^2, b^2 to both sides and squareroot it.
    lal >= a
    lbl >= b
    therefore lal + lbl >= la - bl

    I dont feel very comfortable justifying inequalities, let alone im not very good at it.
    are there any ways to justify without feeling too awkard? some keywords i need to be using?
    please help, i want to learn more
  2. jcsd
  3. Sep 25, 2010 #2


    User Avatar
    Science Advisor

    If you have "proved" it then that is a "justification"?

    Are you asking how to justify each step?

    You first say that [itex]-2ab\le |2ab|[/itex].
    Justify that by looking at cases. If [itex]ab\le 0[/itex] then the two sides are equal. If [itex]ab> 0[/itex] then the left side is negative and the right side is positive.

    Next you have [itex]a^2- 2ab+ b^2\le a^2+ |2ab|+ b^2[/itex] which is true because you have added the same thing to both sides of the inequality.

    Next, [itex](a- b)^2\le (|a|+ |b|)^2[/itex]
    Okay, the left side is exactly the same as the left side in the previous inequality but I would recommend adding something to the previous inequality:
    [itex]a^2- 2ab+ b^2\le a^2+ |2ab|+ b^2= |a|^2+ 2|a||b|+ |b|^2[/itex]
    That last equality is true because [itex]x^2= |x|^2[/itex] for any real number and |xy|= |x||y| for any real numbers.

    Finally, from [itex](a- b)^2\le (|a|+ |b|)^2[/itex]
    you derive [itex]|a- b|\le |a|+ |b|[/itex].

    That is true because if x and y are positive numbers and x> y, then x= y+ a for some positive a so [itex]x^2= y^2+ 2ay+ a^2[/itex] so that [itex]x^2[/itex] is equal to [itex]y^2[/itex] plus some positive number.

    Yes, that is a perfectly valid proof.
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