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Math Help - Proving an Inequality

  1. #1
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    Proving an Inequality

    So i did this last week, and nailed it. The teacher hasn't given it back yet and I was trying to do it again, now I'm stuck.

    I need to prove that

    [\dfrac{1}{2}(a+b)] ^2\leq \dfrac{1}{2}(a^2+b^2)

    I do know that the inequality is equivalent to 0 \leq a^2 -2ab +b^2
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hello,
    Quote Originally Posted by hockey777 View Post
    So i did this last week, and nailed it. The teacher hasn't given it back yet and I was trying to do it again, now I'm stuck.

    I need to prove that

    [\dfrac{1}{2}(a+b)] ^2\leq \dfrac{1}{2}(a^2+b^2)

    I do know that the inequality is equivalent to 0 \leq a^2 -2ab +b^2
    And you also know that a^2-2ab+b^2=(a-b)^2 and that the square of a real number is non negative, don't you ?
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  3. #3
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    Got it, sorry. I was trying to do it a weird way with theorems.
    Last edited by hockey777; September 4th 2008 at 01:00 PM.
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  4. #4
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    It is well known that a^2  + b^2  \ge 2ab.
    Now add to both parts: 2a^2  + 2b^2  \ge a^2  + 2ab + b^2  = \left( {a + b} \right)^2 .
    Regroup:  \frac{1}{2}\left( {a^2  + b^2 } \right) \ge \frac{1}{4}\left( {a + b} \right)^2  = \left[ {\frac{1}{2}\left( {a + b} \right)} \right]^2
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