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Analysis Homework. Proof of Arithmetic-Means Inequality

  1. Feb 8, 2010 #1
    From the Text, Introduction to Analysis, by Arthur Mattuck pg 32 2-3 (a)

    1. The problem statement, all variables and given/known data
    Prove: for any a,b [tex]\geq[/tex] 0, [tex]\sqrt{ab}[/tex] [tex]\leq[/tex] [tex]\frac{\left(a+b\right)}{2}[/tex]
    with equality holding if and only if a =b

    2. Relevant equations

    All Perfect squares are [tex]\geq[/tex] 0

    3. The attempt at a solution

    I wasn't sure where to go with this but I took the inequality:
    [tex]\sqrt{ab}[/tex] [tex]\leq[/tex] [tex]\frac{\left(a+b\right)}{2}[/tex]

    and squared both sides to give me:

    ab [tex]\leq[/tex] [tex]\frac{\left(a+b\right)^{2}}{4}[/tex]

    I then separated the right which gave me:

    ab [tex]\leq[/tex] [tex]\frac{a^{2}}{4}[/tex] + [tex]\frac{ab}{2}[/tex] + [tex]\frac{b^{2}}{4}[/tex]

    I multiplied both sides by 4 which gave me:

    4ab [tex]\leq[/tex] [tex]a^{2}[/tex] + [tex]2ab[/tex] + [tex]b^{2}[/tex]

    I quickly saw that 4ab [tex]\leq[/tex] [tex]\left(a+b\right)^{2}[/tex]

    From the known equation: 0 [tex]\leq[/tex] [tex]\left(a+b\right)^{2}[/tex]

    I added the two together and got: 4ab [tex]\leq[/tex]2 [tex]\left(a+b\right)^{2}[/tex]

    which divided by 4 equals: ab [tex]\leq[/tex] [tex]\frac{\left(a+b\right)^{2}}{2}[/tex]

    This as far as I got. Help would be appreciated.
     
  2. jcsd
  3. Feb 8, 2010 #2

    Mark44

    Staff: Mentor

  4. Feb 8, 2010 #3
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