\[(a+b+c) \left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} \right)\geq 9+3\sqrt[3]{\dfrac{(a-b)^2(b-c)^2(c-a)^2}{a^2b^2c^2}}\]

Prove the inequality above, where \(a,b\) and \(c\) are positive reals.

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## Comments

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TopNewestThe inequality look exactly like this: \[\frac {a}{b} + \frac {b}{a} + \frac {a}{b} + \frac {b}{a} + \frac {c}{a} + \frac {a}{c} + \frac {b}{c} + \frac {c}{b} \geq 3\sqrt[3]{(1 - \frac {b}{a})^{2}*(1- \frac{c}{b})^{2}*(1 - \frac {a}{c})^{2}}+6\] Let \[x=\frac{b}{a}\,\ y=\frac{c}{b}\,\ z=\frac{a}{c}\], then \[xyz=1\] Reforming the inequality: \[x + \frac{1}{x} - 2 + y + \frac{1}{y} - 2+ z + \frac{1}{z} - 2 \geq 3\sqrt[3]{(1 - x)^{2}*(1- y)^{2}*(1 - z)^{2}}\] or \[\frac{(x-1)^2}{x}+\frac{(y-1)^2}{y}+\frac{(z-1)^2}{z} \geq 3\sqrt[3]{(1 - x)^{2}*(1- y)^{2}*(1 - z)^{2}}\] By arithmetic mean - geometric mean : Left side \[\geq 3\sqrt[3]{(1 - x)^{2}*(1- y)^{2}*(1 - z)^{2}*\frac{1}{xyz}}\] and we know that \[xyz=1\] I am very bad with formatting problems. Hope that you will understand:

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This is either an incomplete solution or completely irrelevant. I hope someone else can finish off my work.

We can homogenize the condition by assuming that \(abc= 1\). Then by arithmetic mean - geometric mean:

\[ \begin{eqnarray} && AM [ (a-b)^2 , (b-c)^2 , (c-a)^2 ] \geq GM [ (a-b)^2 , (b-c)^2 , (c-a)^2 ] \\ &&\Rightarrow 3 \sqrt[3]{ (a-b)^2 (b-c)^2 (c-a)^2} \leq 2 (a^2 + b^2 + c^2 -ab-ac-bc) \; . \end{eqnarray} \]

And because \( \dfrac1a + \dfrac1b + \dfrac1c = \dfrac{ab+ac+bc}{abc} = ab + bc + ac\), proving the inequality in question is equivalent to proving the following inequality:

\[ (a + b+c)(ab+ ac+ bc) - 9 \geq 2 (a^2 + b^2 + c^2 -ab-ac-bc) \; .\]

Now I'm stuck. =(

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Hard problem.I had checked the solution by those people from Viet Nam team pre IMO.I don't understand anything !

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Please write out that solution. Maybe it can help me obtain the proper working...

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I've homogenised further and simplified to get what we want to prove as

\[[2, 1, 0] + [\frac{4}{3}, \frac {4}{3}, \frac {4}{3} ] \geq [\frac {7}{3}, \frac {1}{3}, \frac {1}{3}] + 6abc\]

Looks like an application of Schur's.

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