# Prove inequality $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{9\sqrt[3]{abc}}{a+b+c}\geq 6$

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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Welcome to MSE. What are your thoughts and what have you tried so far? Is this homework? If so, go ahead and add the homework tag to the question. Don't worry, people will still help you! – Cameron Buie Aug 22 '12 at 22:13
@CameronBuie are you aware of this meta thread? – user2468 Aug 22 '12 at 22:17
I am, actually. I just didn't remember it until I was nearly done writing the comment. ;-) – Cameron Buie Aug 22 '12 at 22:39

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 $x^2+y^2 \geq 2xy$
Could you explain how you arrived at the first inequality from AM-GM? I'm just getting $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq\frac{3}{\sqrt[3]{abc}}\Rightarrow 1+ \frac{a}{b} +\frac{a}{c}\geq\frac{3a}{\sqrt[3]{abc}}$ – Andrew Aug 22 '12 at 23:16
yes, you have $\frac{a}{b}+\frac{a}{b}+\frac{b}{c} \geq 3 ( \frac{a}{b} \frac{a}{b} \frac{b}{c} )^{\frac{1}{3}} = 3 (\frac{a ^2}{bc}) ^{\frac{1}{3}}= 3 (\frac{a ^2a}{abc}) ^{\frac{1}{3}}= 3a \frac{1}{(abc)^{\frac{1}{3}}}$ is it clear now? – clark Aug 22 '12 at 23:21