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The AM-GM Inequality

  1. Sep 21, 2008 #1
    1. The problem statement, all variables and given/known data
    Show that:
    0 <= A <= B

    [tex] A \leq \sqrt{AB} \leq (A + B)/ 2 \leq B [/tex]


    2. Relevant equations
    root AB = geometric mean
    (A + B)/ 2 = arithmetic mean

    <= means less then or equal to.


    3. The attempt at a solution

    I managed to come up with something for the (A + B)/ 2 part.

    a < = (a+a)/2 <= (a+b)/2 <= (b+b)/2 <= b

    that shows that (a+b)/2 is less then b but greater then a. i can't figure out how to show that with root ab.

    Any help would be great!
     
    Last edited: Sep 21, 2008
  2. jcsd
  3. Sep 21, 2008 #2
    Re: Inequality

    Hint: What do you know about (B - A)^2 ?
     
  4. Sep 21, 2008 #3
    Re: Inequality

    i added something i forgot to mention earlier.

    (B-A)^2 = B^2 -2ab + A^2
    Assuming that (B-A)^2 = 0
    And if i were to rearrange it i'd get B^2 +A^2 = 2ab
    And then if divided the 2 and took the root i'd be left with B+A/root 2 = root AB

    That root 2 is still throwing me off tho. :S
     
  5. Sep 21, 2008 #4
    Re: Inequality

    Why the equality? Since C^2 >= 0 for all C, we know that (B - A)^2 >= 0.
    So as you arranged it, we have B^2 + A^2 >= 2AB
    That is equivalent to A^2 + 2AB + B^2 >= 2AB + 2AB = 4AB.
    Can you see where to go from here?
     
  6. Sep 21, 2008 #5
    Re: Inequality

    :O the light bulb just went on. Thanks a lot :).
     
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