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  1. #1
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    a and b being two strictly positive numbers , compare (a+b)/2 and √a√b
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  2. #2
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    Quote Originally Posted by lebanon View Post
    a and b being two strictly positive numbers , compare (a+b)/2 and √a√b
    Consider (\sqrt{a}-\sqrt{b})^2=(\sqrt{a})^2-2\sqrt{a}\sqrt{b}+(\sqrt{b})^2=a+b-2\sqrt{a}\sqrt{b}.

    Then divide by 2 to get \displaystyle\frac{(\sqrt{a}-\sqrt{b})^2}{2}=\frac{a+b}{2}-\sqrt{a}\sqrt{b}. But since \displaystyle\frac{(\sqrt{a}-\sqrt{b})^2}{2} is nonnegative for any positive a and b (why?) it follows that \displaystyle\frac{a+b}{2}-\sqrt{a}\sqrt{b}\geq0.

    Therefore, \displaystyle\frac{a+b}{2}\geq\sqrt{a}\sqrt{b} for any positive numbers a and b .
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  3. #3
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    And it can further be proven by simple induction that:

    \displaystyle \left(\prod_{1\le i \le n}a_{i}\right)^{\frac{1}{n}} \le \frac{1}{n}\sum_{1\le i\le n}a_{i} for a_{i} \in\mathbb{R}^{+} and n\in\mathbb{N}.
    Last edited by TheCoffeeMachine; January 7th 2011 at 12:24 AM.
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