by v_Enhance, Jul 11, 2011, 12:32 AM 
, 
, 
,![\[ \frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{c+a} + \frac{1}{2\sqrt[3]{abc}} \geq \frac{(a+b+c+\sqrt[3]{abc})^2}{(a+b)(b+c)(c+a)} \]](http://latex.artofproblemsolving.com/3/1/b/31b6043d8d82132e59c4f45f0331293a19fc456b.png)
by
v_Enhance, Jul 11, 2011, 12:37 AM
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