A very hard inequality from Vo Quoc Ba Can

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A very hard inequality from Vo Quoc Ba Can

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Source: Vo Quoc Ba Can

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#1 h Sep 4, 2016, 10:28 AM • 2 Y

Y by Adventure10, Mango247

Let $a,b,c>0$ satisfy $a+b+c=1$ .Prove:
$\frac{36}{a^2b+b^2c+c^2a}+\frac{1}{abc} \ge 343$ equality occur

This post has been edited 4 times. Last edited by thuanz123, Sep 4, 2016, 10:32 AM

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#2 h Sep 4, 2016, 10:37 AM • 2 Y

Y by Adventure10, Mango247

arqady,I know you have a nice solution for this

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sqing

#3 h Sep 4, 2016, 10:53 AM • 2 Y

Y by Adventure10, Mango247

thuanz123 wrote:

Let $a,b,c>0$ satisfy $a+b+c=1$ .Prove:
$\frac{36}{a^2b+b^2c+c^2a}+\frac{1}{abc} \ge 343$ equality occur

http://www.artofproblemsolving.com/community/c6h404630p2256953

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#4 h Sep 4, 2016, 11:09 AM • 2 Y

Y by Adventure10, Mango247

I've already known that solution
I want a nicer solution for this

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#5 h Sep 4, 2016, 11:19 AM • 2 Y

Y by Adventure10, Mango247

a very very ugly solution for this is after full expanding
$\sum{\left(1029-\frac{36(2ab+c^2)}{(\sum{a^2b})^2}-\frac{bc}{a^2b^2c^2}\right)^2}$ will get the answer

This post has been edited 2 times. Last edited by thuanz123, Sep 4, 2016, 11:20 AM

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#6 h Sep 4, 2016, 11:29 AM • 3 Y

Y by zdyzhj, Adventure10, Mango247

another solution is using
$\sum{\left(2(bc+ca-2ab)+a^2+b^2-2c^2\right)^2}\sum{\left(ca+ab-2bc\right)^2} \ge \left(\sum{(2(bc+ca-2ab)+a^2+b^2-2c^2)(ca+ab-2bc)}\right)^2$
will get the answer

This post has been edited 3 times. Last edited by thuanz123, Sep 7, 2016, 3:41 PM

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arqady

#7 h Sep 7, 2016, 6:44 AM • 1 Y

Y by Adventure10

thuanz123 wrote:

another solution is using
$\sum{\left(a^2+b^2-2c^2\right)^2}\sum{\left(bc+ca-2ab\right)^2} \ge \left(\sum{(a^2+b^2-2c^2)(bc+ca-2ab)}\right)^2$
will get the answer

For which problem? Is it a joke?

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#8 h Sep 7, 2016, 3:41 PM • 1 Y

Y by Adventure10

arqady wrote:

thuanz123 wrote:

another solution is using
$\sum{\left(a^2+b^2-2c^2\right)^2}\sum{\left(bc+ca-2ab\right)^2} \ge \left(\sum{(a^2+b^2-2c^2)(bc+ca-2ab)}\right)^2$
will get the answer

For which problem? Is it a joke?

sorry edited wrong typing

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zdyzhj

#9 h Sep 9, 2016, 8:09 AM • 1 Y

Y by Adventure10

thuanz123 wrote:

another solution is using
$\sum{\left(2(bc+ca-2ab)+a^2+b^2-2c^2\right)^2}\sum{\left(ca+ab-2bc\right)^2} \ge \left(\sum{(2(bc+ca-2ab)+a^2+b^2-2c^2)(ca+ab-2bc)}\right)^2$
will get the answer

??????????????????

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#10 h Sep 9, 2016, 3:15 PM • 2 Y

Y by Adventure10, Mango247

zdyzhj wrote:

thuanz123 wrote:

another solution is using
$\sum{\left(2(bc+ca-2ab)+a^2+b^2-2c^2\right)^2}\sum{\left(ca+ab-2bc\right)^2} \ge \left(\sum{(2(bc+ca-2ab)+a^2+b^2-2c^2)(ca+ab-2bc)}\right)^2$
will get the answer

??????????????????

what's up

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zdyzhj

#11 h Sep 11, 2016, 1:03 AM • 2 Y

Y by Adventure10, Mango247

thuanz123 wrote:

zdyzhj wrote:

thuanz123 wrote:

another solution is using
$\sum{\left(2(bc+ca-2ab)+a^2+b^2-2c^2\right)^2}\sum{\left(ca+ab-2bc\right)^2} \ge \left(\sum{(2(bc+ca-2ab)+a^2+b^2-2c^2)(ca+ab-2bc)}\right)^2$
will get the answer

??????????????????

what's up

what is the rqlationship between this one and the orignal one

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#12 h Sep 11, 2016, 4:16 AM • 1 Y

Y by Adventure10

zdyzhj wrote:

what is the rqlationship between this one and the orignal one

if you full expanding and do something then you'll get the original one

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zdyzhj

#13 h Sep 11, 2016, 6:41 AM • 2 Y

Y by Adventure10, Mango247

thuanz123 wrote:

a very very ugly solution for this is after full expanding
$\sum{\left(1029-\frac{36(2ab+c^2)}{(\sum{a^2b})^2}-\frac{bc}{a^2b^2c^2}\right)^2}$ will get the answer

this is symetrical inequality, the orignal one is not,how can you deduce it

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#14 h Sep 11, 2016, 6:46 AM • 2 Y

Y by Adventure10, Mango247

zdyzhj wrote:

thuanz123 wrote:

a very very ugly solution for this is after full expanding
$\sum{\left(1029-\frac{36(2ab+c^2)}{(\sum{a^2b})^2}-\frac{bc}{a^2b^2c^2}\right)^2}$ will get the answer

this is symetrical inequality, the orignal one is not,how can you deduce it

ahhh that's a wrong one I've checked

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