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Proving this inequality

  1. Sep 18, 2009 #1
    1. The problem statement, all variables and given/known data

    Given that 0 <or equal to a <or equal to b, show that

    a <or equal to sqrt(ab) <or equal to (a+b)/2 <or equal to b

    The number sqrt(ab) is called the geometric mean of a and b

    2. Relevant equations

    Not sure -_-

    3. The attempt at a solution

    a = sqrt(aa) <or equal to sqrt(ab) <or equal to sqrt(bb) = b
    a = (a+a)/2 <or equal to (a+b)/2 <or equal to (b+b)/2 = b

    Not sure what to do now -_-
     
  2. jcsd
  3. Sep 18, 2009 #2
    There are 3 things you need to show:

    a < root ab
    root ab < (a+b)/2
    (a+b)/2 < b

    So you have the given condition 0 < a < b (Let's assume that < or > also includes the equal to, since I don't know how to use the notations)

    since 0 < a < b, a < b as well

    try working around a < b

    PS. is this from the MAT137 problem set?
     
  4. Sep 18, 2009 #3
    a <= b;
    a*a <= b*a;
    sqrt(a*a) <= sqrt(b*a)
    a <= sqrt(b*a)

    use similar logic for the other ones.
     
  5. Sep 18, 2009 #4
    I'm not sure how to prove sqrt(ab) <= (a+b)/2

    And yes this is for MAT137 -_-
     
  6. Sep 18, 2009 #5

    Dick

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    Homework Helper

    You know both sides are positive or zero. You can square both sides and the inequality still holds. Can you see the next step?
     
  7. Sep 19, 2009 #6
    I know that b >= a >= 0
    So sqrt(b) >= sqrt(a) >= 0
    I know that b - a >= 0
    So sqrt(b) - sqrt(a) >= 0
    I know that sqrt of any number is >= 0
    So sqrt(sqrt(b) - sqrt(a)) >= 0
    Now I square now sides twice
    (sqrt(b) - sqrt(a))^2 >= 0
    Expand
    b - 2sqrt(ab) + a >= 0
    a + b / 2 >= sqrt(ab)

    I'm pretty sure this is right.. though I only came this part with someone else' help.
    Now I'm wondering if there is any other way to solve this and how I can learn to think this way and be able to solve other similar questions.. any advice would be appreciated, thanks.
     
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