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Thread: Inequality

  1. #1
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    Inequality

    Let $\displaystyle x,y,z \in R$ such that $\displaystyle x^2+y^2+z^2=1$. Prove $\displaystyle |x+y+z| \leq \sqrt{3}$. I thinking along the lines of $\displaystyle AGM \leq GM$.
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  2. #2
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    Re: Inequality

    Quote Originally Posted by brucewayne View Post
    Let $\displaystyle x,y,z \in R$ such that $\displaystyle x^2+y^2+z^2=1$.
    Prove $\displaystyle |x+y+z| \leq \sqrt{3}$.
    Here are the facts you need.
    $\displaystyle 2|xy|\le x^2+y^2$

    $\displaystyle |x+y+z|^2\le (|x|+|y|+|z|)^2=x^2+y^2+z^2+2|xy|+2|xz|+2|yz|$
    Last edited by Plato; Oct 16th 2012 at 07:11 AM.
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    Re: Inequality

    Ok, I am heading in the right direction...

    $\displaystyle 1+2x^2+2y^2+2z^2 \leq \sqrt{3}$
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    Re: Inequality

    Quote Originally Posted by brucewayne View Post
    Ok, I am heading in the right direction...

    $\displaystyle 1+2x^2+2y^2+2z^2 \leq \sqrt{3}$
    Factor out the 2 in the last three terms.
    You have
    $\displaystyle |x+y+z|^2\le 3$
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  5. #5
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    Re: Inequality

    Ok, Maybe I am missing something, but what happened to 1?
    Last edited by brucewayne; Oct 16th 2012 at 08:24 AM.
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    Re: Inequality

    Quote Originally Posted by brucewayne View Post
    Ok, Maybe I am missing something, but what happens to 1?
    Come man do basic mathamatics.
    $\displaystyle 1+2x^2+2y^2+2z^2=1+2(x^2+y^2+z^2)=1+2(1)=3$
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  7. #7
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    Re: Inequality

    Dang it, I really do over analyze things at times. My sincerest apologies; I am trying to get better at this.
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