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Let $a,b,c>0$. What is the proof that: $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{9\sqrt[3]{abc}}{a+b+c}\geq 6$

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Take $ \frac{a}{b}+\frac{a}{b}+\frac{b}{c}\geq 3\frac{a}{ (abc)^\frac{1}{3}}$ $ \frac{b}{c}+\frac{b}{c}+\frac{c}{a}\geq 3\frac{b}{ (abc)^\frac{1}{3}}$ $ \frac{c}{a}+\frac{c}{a}+\frac{a}{b}\geq 3\frac{c}{ (abc)^\frac{1}{3}}$ from AM-GM and then add them and you get $\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \geq \frac{a+b+c}{(abc)^\frac{1}{3}}$ So it suffices to prove that $\left ( \frac{(a+b+c)}{(abc)^\frac{1}{3}}\, \, \, \, \, \, +\frac{9 (abc)^\frac{1}{3}}{a+b+c}\, \, \, \, \,\right )\geq 6$ which holds from the basic inequality $x^2+y^2 \geq 2xy$

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    Nice work! (+1)2012-08-24