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Ovchinnikov Denis

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Ovchinnikov Denis

#1 h Sep 8, 2010, 12:20 PM • 2 Y

Y by Adventure10, Mango247

For positive $a,b,c$ , such that $abc=1$ prove the inequality:
$4(\sqrt[3]{\frac{a}{b}}+\sqrt[3]{\frac{b}{c}}+\sqrt[3]{\frac{c}{a}}) \leq 3(2+a+b+c+\frac{1}{a}+\frac{1}{b}+ \frac{1}{c})^{\frac{2}{3}}$ .

This post has been edited 1 time. Last edited by Ovchinnikov Denis, Sep 8, 2010, 12:59 PM

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kuing

#2 h Sep 8, 2010, 12:47 PM • 2 Y

Y by Adventure10, Mango247

Ovchinnikov Denis wrote:

For positive $a,b,c$ , such that $abc=1$ prove the inequality:
$4(\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{c}}+\sqrt{\frac{c}{a}}) \leq 3(2+a+b+c+\frac{1}{a}+\frac{1}{b}+ \frac{1}{c})^{\frac{2}{3}}$ .

try $a=1,b=4,c=\frac{1}{4}$ .

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Ovchinnikov Denis

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Ovchinnikov Denis

#3 h Sep 8, 2010, 12:57 PM • 1 Y

Y by Adventure10

kuing wrote:

Ovchinnikov Denis wrote:

For positive $a,b,c$ , such that $abc=1$ prove the inequality:
$4(\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{c}}+\sqrt{\frac{c}{a}}) \leq 3(2+a+b+c+\frac{1}{a}+\frac{1}{b}+ \frac{1}{c})^{\frac{2}{3}}$ .

try $a=1,b=4,c=\frac{1}{4}$ .

Sorry, in left side cubic squares

Corrected

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kuing

#4 h Sep 8, 2010, 1:18 PM • 2 Y

Y by dizzy, Adventure10

Ovchinnikov Denis wrote:

For positive $a,b,c$ , such that $abc=1$ prove the inequality:
$4(\sqrt[3]{\frac{a}{b}}+\sqrt[3]{\frac{b}{c}}+\sqrt[3]{\frac{c}{a}}) \leq 3(2+a+b+c+\frac{1}{a}+\frac{1}{b}+ \frac{1}{c})^{\frac{2}{3}}$ .

let $a = \frac{y}{x},b = \frac{z}{y},c = \frac{x}{z},x,y,z > 0$ , then

$\Leftrightarrow 4\left( {\sqrt[3]{{\frac{{y^2 }}{{zx}}}} + \sqrt[3]{{\frac{{z^2 }}{{xy}}}} + \sqrt[3]{{\frac{{x^2 }}{{yz}}}}} \right) \le 3\left( {2 + \frac{x}{y} + \frac{y}{z} + \frac{z}{x} + \frac{y}{x} + \frac{z}{y} + \frac{x}{z}} \right)^{\frac{2}{3}}$

$\Leftrightarrow 4\left( {\frac{{x + y + z}}{{\sqrt[3]{{xyz}}}}} \right) \le 3\left( {\frac{{\left( {x + y} \right)\left( {y + z} \right)\left( {z + x} \right)}}{{xyz}}} \right)^{\frac{2}{3}}$

$\Leftrightarrow 64xyz\left( {x + y + z} \right)^3 \le 27\left( {x + y} \right)^2 \left( {y + z} \right)^2 \left( {z + x} \right)^2$

$\Leftrightarrow xyz\left( {\frac{{x + y + z}}{3}} \right)^3 \le \left( {\frac{{\left( {x + y} \right)\left( {y + z} \right)\left( {z + x} \right)}}{8}} \right)^2,$

by $\frac{{\left( {x + y + z} \right)\left( {xy + yz + zx} \right)}}{9} \le \frac{{\left( {x + y} \right)\left( {y + z} \right)\left( {z + x} \right)}}{8}$ , so we just need to prove

$xyz\left( {\frac{{x + y + z}}{3}} \right)^3 \le \left( {\frac{{\left( {x + y + z} \right)\left( {xy + yz + zx} \right)}}{9}} \right)^2$

$\Leftrightarrow 3xyz\left( {x + y + z} \right) \le \left( {xy + yz + zx} \right)^2,$

obvious true, done.

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#5 h Sep 9, 2010, 10:55 AM • 1 Y

Y by Adventure10

2006丝绸之路不等式

Attachments:

2006年丝绸之路.pdf (54kb)

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arqady

#6 h Jul 24, 2012, 12:28 PM • 2 Y

Y by Adventure10, Mango247

Ovchinnikov Denis wrote:

For positive $a,b,c$ , such that $abc=1$ prove the inequality:
$4(\sqrt[3]{\frac{a}{b}}+\sqrt[3]{\frac{b}{c}}+\sqrt[3]{\frac{c}{a}}) \leq 3(2+a+b+c+\frac{1}{a}+\frac{1}{b}+ \frac{1}{c})^{\frac{2}{3}}$ .

See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=594965#p594965

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sqing

#7 h Dec 22, 2013, 2:11 AM • 2 Y

Y by Adventure10, Mango247

See also
5th Silk Road Mathematical Competition SRMC2006

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