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Need help with proving something?

  1. Sep 13, 2009 #1
    1. The problem statement, all variables and given/known data

    0 < a < b

    Must prove that a < sqrt(ab) < (a+b)/2 < b

    2. Relevant equations

    Well, relevant info. I am using Spivak's calculus book, where we have to prove everything.

    3. The attempt at a solution

    I'm not sure if I'm doing this right but here's my start.

    I'll start with a < b. Multiply both sides by a, and you get a times a < a times b.

    Then you get a^2 < ab

    Then square root of a^2 < square root of ab.

    a < square root of ab.

    So I think i have the first part of that, but I'm not sure where to start for a+b/2?
     
  2. jcsd
  3. Sep 13, 2009 #2

    VietDao29

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    Homework Helper

    [tex]\frac{a + b}{2} \geq \sqrt{ab}[/tex]

    This is actually a special case of http://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means#The_inequality" (the case for n = 2, 2 terms). This case (n = 2) can be easily proven by isolating everything to one side, and use the fact that:

    [tex](x - y) ^ 2 = x ^ 2 - 2xy + y ^ 2 \geq 0[/tex]

    And for [tex]x \neq y[/tex], we have the equality:

    [tex](x - y) ^ 2 = x ^ 2 - 2xy + y ^ 2 {\color{red}>} 0[/tex]

    Let's see if you can take it from here.

    And the last equality should be easy. :)
     
    Last edited by a moderator: Apr 24, 2017
  4. Sep 13, 2009 #3
    VietDao29 did you mean:

    [tex](\sqrt{x}-\sqrt{y})^2 \geq 0 [/tex] ?
     
  5. Sep 13, 2009 #4

    VietDao29

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    No, I mean x, y; x, y in general.
     
  6. Sep 13, 2009 #5
    My post got deleted?!! Im new to the forum here.. someone please shed light.
     
  7. Sep 14, 2009 #6
    If he use x=a, and y=b in the inequality that I posted, it will be very easy to prove it. :smile:
     
  8. Sep 14, 2009 #7

    HallsofIvy

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    You gave a complete solution rather than helping hints.
     
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