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# Another conservation problem

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

Prove the inequality above, where $$a,b$$ and $$c$$ are positive reals.

Note by Ms Ht
8 months, 1 week ago

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The 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: · 8 months ago

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. =( · 8 months ago

Hard problem.I had checked the solution by those people from Viet Nam team pre IMO.I don't understand anything ! · 8 months ago

Please write out that solution. Maybe it can help me obtain the proper working... · 8 months ago

Ok · 8 months ago

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. · 8 months ago