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#1

$Old$ November 6th, 2009, 03:12 PM

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For any $a, b, c > 0$ with $abc = 1$ , prove that

$\frac{1}{a^3(b+c)}+\frac{1}{b^3(c+a)}+\frac{1}{c^3(a+b)} \ge \frac{3}{2}$

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#2

$Old$ November 6th, 2009, 03:38 PM

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with abc = 1
then $a=1\ b=1 \ and \ c=1$

therefore

$\frac{1}{1^3(1+1)} + \frac{1}{1^3(1+1)} + \frac{1}{1^3(1+1)} \geq \frac{3}{2}$

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#3

$Old$ November 6th, 2009, 03:59 PM

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you don't have a proof, that's a particular case you just took.

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#4

$Old$ November 10th, 2009, 08:05 AM

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Originally Posted by bigwave $View Post$

with abc = 1
then $a=1\ b=1 \ and \ c=1$

therefore

$\frac{1}{1^3(1+1)} + \frac{1}{1^3(1+1)} + \frac{1}{1^3(1+1)} \geq \frac{3}{2}$

If it only was that easy!

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