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May 11th, 2015, 11:44 PM   #1
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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.
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May 12th, 2015, 02:22 AM   #2
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By the AM-GM inequality, $\dfrac1a + \dfrac1b + \dfrac1c \geqslant \dfrac{3}{\sqrt[3]{abc}}$.

By the AM-GM inequality, $\sqrt[3]{abc} + \dfrac{3}{\sqrt[3]{abc}} \geqslant 2\sqrt3$.
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May 12th, 2015, 02:36 AM   #3
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Math Focus: Abstract algebra
Another method – 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 AM–GM to $\frac1{\sqrt a}+\frac1{\sqrt b}+\frac1{\sqrt b}\geqslant\frac3{(abc)^{1/6}}$.
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May 12th, 2015, 02:56 AM   #4
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Thanks for both of your solutions, but I think skipjack's solution is neater and simpler.
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