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Math Help - request help with proof

  1. #1
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    request help with proof

    Show that (a + b)/2 is greater than or equal to squareroot (ab) for 0 < a ≤ b.


    I'm in a class in which I'm being exposed to proofs for the first time, and I'm quite unsure about where to begin.

    I can input examples to show that it's true, but that's obviously not what I need to do here.

    While I would appreciate a complete explanation, even a gentle nudge in the right direction would be helpful.

    Thanks!
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  2. #2
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    Quote Originally Posted by thejollywag View Post
    Show that (a + b)/2 is greater than or equal to squareroot (ab) for 0 < a ≤ b.


    I'm in a class in which I'm being exposed to proofs for the first time, and I'm quite unsure about where to begin.

    I can input examples to show that it's true, but that's obviously not what I need to do here.

    While I would appreciate a complete explanation, even a gentle nudge in the right direction would be helpful.

    Thanks!
    haven't done this in awhile, but if I remember correctly, a proof may look somewhat like this

    use the general idea that if x \ge y , then x - y \ge 0

    using the two values in question ...

    \frac{a+b}{2} - \sqrt{ab} =

    \frac{a+b}{2} - \frac{2\sqrt{ab}}{2} =

    \frac{a - 2\sqrt{ab} + b}{2} =

    \frac{(\sqrt{a} - \sqrt{b})^2}{2} \ge 0
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  3. #3
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    I appreciate it, Skeeter. That's an extremely helpful explanation.
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