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Indirect Proof

  1. Apr 11, 2008 #1
    1. The problem statement, all variables and given/known data

    show that sqrt(x+b) - sqrt(x-a) >= sqrt(x+a) - sqrt(x-b)
    where x >= a >= b >= 0

    2. Relevant equations

    none

    3. The attempt at a solution

    My instructor said that we had to use an indirect proof.
    The give statement is "if x >= a >= b >= 0, then sqrt(x+b) - sqrt(x-a) >= sqrt(x+a) - sqrt(x-b)"
    So for the proof, the statement is "if x >= a >= b >= 0, then sqrt(x+b) - sqrt(x-a) < sqrt(x+a) - sqrt(x-b)"

    By manipulating sqrt(x+b) - sqrt(x-a) < sqrt(x+a) - sqrt(x-b), I get abs(a) < abs(b).

    Heres the algebra
    sqrt(x+b) - sqrt(x-a) < sqrt(x+a) - sqrt(x-b)
    sqrt(x+b) + sqrt(x-b) < sqrt(x+a) + sqrt(x-a)
    x + b + 2 sqrt(x^2 - b^2) + x - b < x + a + 2 sqrt(x^2 - a^2) + x - a
    2x + 2 sqrt(x^2 - b^2) < 2x + 2 sqrt(x^2 - a^2)
    sqrt(x^2 - b^2) < sqrt(x^2 - a^2)
    x^2 - b^2 < x^2 - a^2
    a^2 < b^2
    abs(a) < abs(b)

    abs(a) < abs(b) contradicts the given: x >= a >= b >= 0
    So my question is does that automatically make "if x >= a >= b >= 0, then sqrt(x+b) - sqrt(x-a) >= sqrt(x+a) - sqrt(x-b)" true? or did I miss something?
     
  2. jcsd
  3. Apr 12, 2008 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    If assuming sqrt(x+b) - sqrt(x-a) < sqrt(x+a) - sqrt(x-b), leads to a contradictionthen it can't be true. If it can't be true what is true?
     
  4. Apr 12, 2008 #3

    tiny-tim

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    Science Advisor
    Homework Helper

    Hi asura! :smile:

    (feel free to copy √ and ² and ≤, and anything else you like, for future use :smile:)

    Personally, I think "inequations" (with < or ≤) are really confusing. :confused:

    Hint: it's often clearer to replace the < by a minus, so you get:

    [√(x+b) + √(x-b)]² - [√(x+a) + √(x-a)]²
    = …
    = 2√(x² - b²) - 2√(x² - a²),
    and carry on from there (starting with "But b² ≤ a²", so …). :smile:
     
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