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tunganh
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Post Posted: May 04, 2007, 10:03 pm • # 1 


Let a,b,c be positive reals such that a+b+c=1. Prove that:
7(ab+bc+ca) \le 9abc+2
 
 
N.T.TUAN
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Post Posted: May 04, 2007, 10:20 pm • # 2 


solution

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tunganh
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Post Posted: May 04, 2007, 10:53 pm • # 3 


Nice solution :)
Can anyone find other solutions to this problems? :D


Last edited by tunganh on May 04, 2007, 11:27 pm, edited 1 time in total.
 
 
nayel
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Post Posted: May 04, 2007, 11:17 pm • # 4 


another solution but not very nice

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tunganh
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Post Posted: May 04, 2007, 11:21 pm • # 5 


nayel wrote:
another solution but not very nice

That's right. It's one of my solutions :D
But I denote: a= \frac{x}{x+y+z}, b= \frac{y}{x+y+z}, c= \frac{z}{x+y+z} :)
 
 
tunganh
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Post Posted: May 04, 2007, 11:41 pm • # 6 


A moment ago, I suggested using:
(a+1)(b+1)(c+1) \ge 8(1-a)(1-b)(1-c)
But I deleted it because I believed that it's useless. And now I find that I miscalculated and it's right :)
It's a nice solution, isn't it?
 
 
tjhance
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Post Posted: May 05, 2007, 6:01 am • # 7 


N.T.TUAN wrote:
We know that (a+b-c)(a-b+c)(-a+b+c)\leq abc


Why is that?

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tunganh
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Post Posted: May 05, 2007, 6:22 am • # 8 


tjhance wrote:
N.T.TUAN wrote:
We know that (a+b-c)(a-b+c)(-a+b+c)\leq abc


Why is that?

Denote a+b-c=x, a-b+c=y,-a+b+c=z
Then the inequality becomes xyz \le \frac{(x+y)(y+z)(z+x)}{8} which is clearly true by AM-GM :D

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I'm tired of being what you want me to be; feeling so faithless, lost under the surface; I don't know what you're expecting of me; put under the pressure, of walking in your shoes; (caught in the undertow, just caught in the undertow); every step that I t
 
 
arqady
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Post Posted: May 05, 2007, 6:38 am • # 9 


tunganh wrote:
tjhance wrote:
N.T.TUAN wrote:
We know that (a+b-c)(a-b+c)(-a+b+c)\leq abc


Why is that?

Denote a+b-c=x, a-b+c=y,-a+b+c=z
Then the inequality becomes xyz \le \frac{(x+y)(y+z)(z+x)}{8} which is clearly true by AM-GM :D

Your proof is wrong. :( Why x, y and z are non-negative numbers? :wink:

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tjhance
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Post Posted: May 05, 2007, 6:41 am • # 10 


tunganh wrote:
tjhance wrote:
N.T.TUAN wrote:
We know that (a+b-c)(a-b+c)(-a+b+c)\leq abc


Why is that?

Denote a+b-c=x, a-b+c=y,-a+b+c=z
Then the inequality becomes xyz \le \frac{(x+y)(y+z)(z+x)}{8} which is clearly true by AM-GM :D


ack i'm sorry but I'm still not seeing it...

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Um. :ninja:
 
 
pohoatza
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Post Posted: May 05, 2007, 6:43 am • # 11 


tunganh wrote:
Let a,b,c be positive reals such that a+b+c=1. Prove that:
7(ab+bc+ca) \le 9abc+2


This is just Schur: 7(ab+bc+ca)(a+b+c) \leq 9abc+2(a+b+c)^{3}.

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tunganh
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Post Posted: May 05, 2007, 7:05 am • # 12 


arqady wrote:
tunganh wrote:
tjhance wrote:
N.T.TUAN wrote:
We know that (a+b-c)(a-b+c)(-a+b+c)\leq abc


Why is that?

Denote a+b-c=x, a-b+c=y,-a+b+c=z
Then the inequality becomes xyz \le \frac{(x+y)(y+z)(z+x)}{8} which is clearly true by AM-GM :D

Your proof is wrong. :( Why x, y and z are non-negative numbers? :wink:

I'm sorry :)
But the inequality just follow Schur inequality :D
 
 
vietkhoa
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Post Posted: May 05, 2007, 7:24 am • # 13 


We easily have (a+b-c)(a-b+c)=[a-(b-c)][a+(b-c)]=a^{2}-(b-c)^{2}\leq a^{2}
Similarity we have three inequalities and when we multiply them, we can complete the proof.
This is the short form of Schur, I think
 
 
N.T.TUAN
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Post Posted: May 05, 2007, 9:39 am • # 14 


We also can use following assertion: Let P(x,y,z) be a symmetric polynomial , then a) is equivalent to b), where
a)P(x,y,z)\geq 0\forall x,y,z\geq 0
b)P(1,1,1)\geq 0,P(1,1,0)\geq 0, P(1,0,0)\geq 0.

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HTA
Riemann Hypothesis

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Post Posted: May 05, 2007, 6:44 pm • # 15 


the above inequalities equivalent to
f(ab)=ab(7-9c)+7c(1-c)-2 \leq 0
because 0 \leq ab \leq \frac{(1-c)^{2}}{4} then all we need to do is to prove that f(0);f(\frac{(1-c)^{2}}{4}\leq 0
f(0)=-7(c-\frac{1}{2})^{2}-\frac{1}{4}\leq 0
and f(\frac{(1-c)^{2}}{4}) =-\frac{(c+1)(3c-1)^{2}}{4}\leq 0
so we are done :)
 
 
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