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 May 11th, 2015, 11:44 PM #1 Newbie   Joined: Dec 2014 From: Singapore Posts: 23 Thanks: 0 Inequality problem Hi guys, here is a problem I'm stuck on: If $a$, $b$ and $c$ are positive reals, prove that $\sqrt[3]{abc}+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\geq2 \sqrt{3}$ I found out if $a=b=c$, then the LHS equals $a+\dfrac{3}{a}$. This is always $\geq2 \sqrt{3}$. But how to show the minimum is at $a=b=c$? Thanks in advance for any solutions.
 May 12th, 2015, 02:22 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,469 Thanks: 2038 By the AM-GM inequality, $\dfrac1a + \dfrac1b + \dfrac1c \geqslant \dfrac{3}{\sqrt[3]{abc}}$. By the AM-GM inequality, $\sqrt[3]{abc} + \dfrac{3}{\sqrt[3]{abc}} \geqslant 2\sqrt3$. Thanks from Olinguito
 May 12th, 2015, 02:36 AM #3 Senior Member     Joined: Apr 2014 From: Greater London, England, UK Posts: 320 Thanks: 156 Math Focus: Abstract algebra Another method â€“ start with: $\displaystyle \left[(abc)^{1/6}-\sqrt{\frac3a}\right]^2+\left[(abc)^{1/6}-\sqrt{\frac3b}\right]^2+\left[(abc)^{1/6}-\sqrt{\frac3c}\right]^2\ \geqslant\ 0$Expanding and rearranging: $\displaystyle 3\sqrt[3]{abc} + 3\left(\frac1a+\frac1b+\frac1c\right)\ \geqslant\ 2\sqrt3(abc)^{1/6}\left(\frac1{\sqrt a}+\frac1{\sqrt b}+\frac1{\sqrt b}\right)$ Now apply AMâ€“GM to $\frac1{\sqrt a}+\frac1{\sqrt b}+\frac1{\sqrt b}\geqslant\frac3{(abc)^{1/6}}$.
 May 12th, 2015, 02:56 AM #4 Newbie   Joined: Dec 2014 From: Singapore Posts: 23 Thanks: 0 Thanks for both of your solutions, but I think skipjack's solution is neater and simpler.

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