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Math Help - Inequality

  1. #1
    Member great_math's Avatar
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    Inequality

    If x,y and z are non-negative reals prove that:

    \frac{\frac{x \sqrt y + y \sqrt z + z \sqrt x}{3} + \frac{y \sqrt x + z \sqrt y + x \sqrt z}{3}}{2} \le \sqrt{\left(\frac{x+y}{2}\right) \left( \frac{y+z}{2} \right) \left(\frac{z+x}{2}\right)}
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  2. #2
    Member Greengoblin's Avatar
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    Looks like you could reduce it to the AM-GM inequality.
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  3. #3
    Moo
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    Quote Originally Posted by Greengoblin View Post
    Looks like you could reduce it to the AM-GM inequality.
    AM-GM inequality would give an inequality in the other direction, namely \geq
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  4. #4
    Member Greengoblin's Avatar
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    oh. oops! I only just glanced at this, sorry
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  5. #5
    Member Greengoblin's Avatar
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    Quote Originally Posted by great_math View Post
    If x,y and z are non-negative reals prove that:

    \frac{\frac{x \sqrt y + y \sqrt z + z \sqrt x}{3} + \frac{y \sqrt x + z \sqrt y + x \sqrt z}{3}}{2} \le \sqrt{\left(\frac{x+y}{2}\right) \left( \frac{y+z}{2} \right) \left(\frac{z+x}{2}\right)}
    I think you jut have to simplyfy both sides down, like so:

    <br />
\frac{x\sqrt{y}+y\sqrt{z}+z\sqrt{x}+y\sqrt{x}+z\sq  rt{y}+x\sqrt{z}}{6}\le \sqrt{\frac{(x+y)(y+z)(z+x)}{2^3}}<br />
    <br />
\frac{x\sqrt{y}+y\sqrt{z}+z\sqrt{x}+y\sqrt{x}+z\sq  rt{y}+x\sqrt{z}}{6}\le \sqrt{\frac{(y^2+xy+xz+yz)(z+x)}{8}}
    <br />
\le \sqrt{\frac{zy^2+xy^2+xyz+x^2y+xz^2+x^2z+yz^2+xyz}  {8}}<br />

    then try and square both sides, and reduce both numerators as much as possible. My guess is they will turn out the same (but I can't be bothered to do it all), so you will have something like \frac{u}{36}\le\frac{u}{8} where u is an expression in x,y, and z.
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