1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    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!

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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook