Another conservation problem

(a+b+c)(1a+1b+1c)≥9+3(a−b)2(b−c)2(c−a)2a2b2c23(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,ba,b and cc are positive reals.

#Algebra

Note by Son Nguyen
5 years ago

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Comments

The inequality look exactly like this: ab+ba+ab+ba+ca+ac+bc+cb≥3(1−ba)2∗(1−cb)2∗(1−ac)23+6\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=ba  y=cb  z=acx=\frac{b}{a}\,\ y=\frac{c}{b}\,\ z=\frac{a}{c}, then xyz=1xyz=1 Reforming the inequality: x+1x−2+y+1y−2+z+1z−2≥3(1−x)2∗(1−y)2∗(1−z)23x + \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 (x−1)2x+(y−1)2y+(z−1)2z≥3(1−x)2∗(1−y)2∗(1−z)23\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 ≥3(1−x)2∗(1−y)2∗(1−z)2∗1xyz3\geq 3\sqrt[3]{(1 - x)^{2}*(1- y)^{2}*(1 - z)^{2}*\frac{1}{xyz}} and we know that xyz=1xyz=1 I am very bad with formatting problems. Hope that you will understand:

HIEU TRAN - 5 years 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=1abc= 1. Then by arithmetic mean - geometric mean:

AM[(a−b)2,(b−c)2,(c−a)2]≥GM[(a−b)2,(b−c)2,(c−a)2]⇒3(a−b)2(b−c)2(c−a)23≤2(a2+b2+c2−ab−ac−bc)  . \begin{aligned} && 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{aligned}

And because 1a+1b+1c=ab+ac+bcabc=ab+bc+ac \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≥2(a2+b2+c2−ab−ac−bc)  . (a + b+c)(ab+ ac+ bc) - 9 \geq 2 (a^2 + b^2 + c^2 -ab-ac-bc) \; .

Now I'm stuck. =(

Pi Han Goh - 5 years ago

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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 !

Son Nguyen - 5 years ago

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

Pi Han Goh - 5 years ago

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@Pi Han Goh – Ok

Son Nguyen - 5 years ago

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

[2,1,0]+[43,43,43]≥[73,13,13]+6abc[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.

Sharky Kesa - 5 years ago
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