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Inquality ...

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Cauchy Inequality

inequalities unsolved

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#1 h Nov 9, 2010, 12:31 PM • 2 Y

Y by Adventure10, Mango247

a , b , c are positive reals

Prove that :

Image not found

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#2 h Nov 10, 2010, 2:31 AM • 1 Y

Y by Adventure10

Wajih wrote:

a , b , c are positive reals

Prove that :

Image not found

by Cauchy inequality,we got $a^{3}c+b^{3}a+c^{3}b\ge a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}$
is true by Muirhead

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arqady

#3 h Nov 10, 2010, 4:04 AM • 1 Y

Y by Adventure10

Wajih wrote:

a , b , c are positive reals

Prove that :
$\frac{a^4}{a^2+ab+b^2}+\frac{b^4}{b^2+bc+c^2}+\frac{c^4}{c^2+ca+a^2}\geq\frac{a^3+b^3+c^3}{a+b+c}$

After expanding we need to prove that
$\sum_{cyc}(a^6c^3+a^5c^4+a^6b^2c+a^6c^2b+a^5c^3b-a^5b^3c-2a^4b^3c^2-a^4c^3b^2-a^3b^3c^3)\geq0$
which is true by AM-GM.
For example: $\sum_{cyc}\left(\frac{3}{4}a^6b^2c+\frac{1}{4}a^6c^2b\right)=\sum_{cyc}\left(\frac{3}{4}a^6b^2c+\frac{1}{4}b^6a^2c\right)\geq\sum_{cyc}a^5b^3c$ ,...

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kuing

#4 h Nov 10, 2010, 4:09 AM • 1 Y

Y by Adventure10

pxchg1200 wrote:

Wajih wrote:

a , b , c are positive reals

Prove that :

Image not found

by Cauchy inequality,we got $a^{3}c+b^{3}a+c^{3}b\ge a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}$
is true by Muirhead

I don't think so, Muirhead need sym, not cyc...
in fact, try $a=2,b=1,c=\frac{1}{4}$

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#5 h Nov 10, 2010, 4:14 AM • 1 Y

Y by Adventure10

kuing wrote:

pxchg1200 wrote:

Wajih wrote:

a , b , c are positive reals

Prove that :

Image not found

by Cauchy inequality,we got $a^{3}c+b^{3}a+c^{3}b\ge a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}$
is true by Muirhead

I don't think so, Muirhead need sym, not cyc...
in fact, try $a=2,b=1,c=\frac{1}{4}$

yes,I made a mistake,you're right kuing!

Can you show me another solution without expanding it?

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kuing

#6 h Nov 11, 2010, 10:38 AM • 2 Y

Y by Adventure10, Mango247

it's hard for me, I'm also waiting a nice solution...

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#7 h Nov 12, 2010, 2:10 AM • 3 Y

Y by Gluncho, Adventure10, Mango247

$\Longleftrightarrow \sum_{cyc}{\left(\frac{a^4}{a^2+ab+b^2}+ab-a^2\right)}\geq \frac{3abc}{a+b+c}$
$\Longleftrightarrow \sum_{cyc}{\frac{ab^3}{a^2+ab+b^2}}\geq \frac{3abc}{a+b+c}$
$\Longleftrightarrow \sum_{cyc}{\frac{b^2}{1+\frac{a}{b}+\frac{b}{a}}}\geq \frac{3abc}{a+b+c}$
By cauchy we have
$\sum_{cyc}{\frac{b^2}{1+\frac{a}{b}+\frac{b}{a}}}\geq \frac{(a+b+c)^2}{3+\frac{a}{b}+\frac{b}{a}+\frac{a}{c}+\frac{c}{a}+\frac{b}{c}+\frac{c}{b}}=\frac{abc(a+b+c)}{ab+bc+ca}$
It suffice to show that
$\frac{abc(a+b+c)}{ab+bc+ca}\geq \frac{3abc}{a+b+c}$
$\Longleftrightarrow (a+b+c)^2\geq 3(ab+bc+ca)$
which is trival.

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arqady

#8 h Nov 12, 2010, 4:51 AM • 1 Y

Y by Adventure10

Great solution, Tourish! Thank you!

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kuing

#9 h Nov 12, 2010, 5:15 AM • 1 Y

Y by Adventure10

indeed, very great!

many thanks

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#10 h Nov 12, 2010, 12:56 PM • 2 Y

Y by Adventure10, Mango247

Thank you very much for solving it ...

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#11 h Nov 12, 2010, 2:08 PM • 2 Y

Y by Adventure10, Mango247

So great solution! thank you very much!

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