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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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  • $\begingroup$ Could you explain how you arrived at the first inequality from AM-GM? I'm just getting $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq\frac{3}{\sqrt[3]{abc}}\Rightarrow 1+ \frac{a}{b} +\frac{a}{c}\geq\frac{3a}{\sqrt[3]{abc}}$ $\endgroup$ – Andrew Aug 22 '12 at 23:16
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    $\begingroup$ 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? $\endgroup$ – clark Aug 22 '12 at 23:21
  • $\begingroup$ Yes, that's awesome:) Thanks:) $\endgroup$ – Andrew Aug 22 '12 at 23:25
  • $\begingroup$ Glad I could help:) $\endgroup$ – clark Aug 22 '12 at 23:37
  • $\begingroup$ Nice work! (+1) $\endgroup$ – user 1357113 Aug 24 '12 at 9:44

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