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Math Help - Inequality proof

  1. #1
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    Inequality proof

    Prove [1/2(a+b)]^2 < or equal to 1/2(a^2+b^2). where a, b are real numbers.

    Can't figure out how to end up with a contradiction or if thats even how to approach the problem.
    Last edited by hayter221; October 5th 2008 at 12:49 PM.
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  2. #2
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    It is well known that a^2  + b^2  \geqslant 2ab.
    It is key to doing this problem.
    \left[ {\frac{{a + b}}{2}} \right]^2  = \frac{{a^2  + {\color{blue}2ab} + b^2 }}{4} \leqslant \frac{{2a^2  + 2b^2 }}{4} = \frac{{a^2  + b^2 }}{2}
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  3. #3
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    Is that some kind of theorem though, our professor asks us to be very specific. We cannot assume anything unless it was proven in our book or we prove it oursleves.
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  4. #4
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    It is well known that a^2  + b^2  \geqslant 2ab.
    PROOF:
    \begin{array}{rcl}<br />
   {\left( {a - b} \right)^2 } &  \geqslant  & 0  \\<br />
   {a^2  - 2ab + b^2 } &  \geqslant  & 0  \\<br />
   {a^2  + b^2 } &  \geqslant  & {2ab}  \\ \end{array}
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  5. #5
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    wow that makes a lot of sense...thanks!
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