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Topic: Interesting inequality (Read 1223 times) 

wonderful
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Interesting inequality
« on: Jun 30^{th}, 2008, 8:18pm » 
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Can you generalize the result? Have A Great Day!

« Last Edit: Jul 1^{st}, 2008, 1:25pm by wonderful » 
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towr
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Re: Interesting inequality
« Reply #1 on: Jul 1^{st}, 2008, 12:54am » 
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Shouldn't the second term have z^{2}+2xz in the numerator ?


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ThudnBlunder
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Re: Interesting inequality
« Reply #2 on: Jul 1^{st}, 2008, 5:54am » 
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on Jul 1^{st}, 2008, 12:54am, towr wrote:Shouldn't the second term have z^{2}+2xz in the numerator ? 
 Undoubtedly. If so, I get f(x,y,z) > 3, but what do I know?


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pex
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Re: Interesting inequality
« Reply #3 on: Jul 1^{st}, 2008, 6:55am » 
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on Jul 1^{st}, 2008, 5:54am, ThudanBlunder wrote: Undoubtedly. If so, I get f(x,y,z) > 3, but what do I know? 
 If we write f(x, y, z) = (y^{2} + 2yz) / (y  z)^{2} + (z^{2} + 2xz) / (x  z)^{2} + (x^{2} + 2xy) / (x  y)^{2}, then isn't lim(m > infinity) f(1, m, m^{2}) = 1?


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towr
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Re: Interesting inequality
« Reply #4 on: Jul 1^{st}, 2008, 7:07am » 
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on Jul 1^{st}, 2008, 6:55am, pex wrote:isn't lim(m > infinity) f(1, m, m^{2}) = 1? 
 That's what I was gonna say (well almost). Maybe we're supposed to assume x,y,z are integers?


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pex
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Re: Interesting inequality
« Reply #5 on: Jul 1^{st}, 2008, 7:09am » 
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on Jul 1^{st}, 2008, 7:07am, towr wrote:Maybe we're supposed to assume x,y,z are integers? 
 Wouldn't the same counterexample still work? Edit: I realize you probably derived the counterexample the same way I did, using (1/m, 1, m). However, f is homogeneous of degree zero...

« Last Edit: Jul 1^{st}, 2008, 7:10am by pex » 
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towr
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Re: Interesting inequality
« Reply #6 on: Jul 1^{st}, 2008, 7:19am » 
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on Jul 1^{st}, 2008, 7:09am, pex wrote:Wouldn't the same counterexample still work? 
 Err, ahum, yes.. Quote:Edit: I realize you probably derived the counterexample the same way I did, using (1/m, 1, m). However, f is homogeneous of degree zero... 
 Actually, I used f(1/m^{2}, 1/m, 1) Or rather, I picked two of them to be practically 0 (but of a different order)


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ThudnBlunder
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Re: Interesting inequality
« Reply #7 on: Jul 1^{st}, 2008, 9:41am » 
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Can't we also say as n > 0, m > infinity then f(n, m, m^{2}) > 0? And as m > infinity, f(m, m+1, z) > infinity. Hence the expression can take all positive values.


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wonderful
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Re: Interesting inequality
« Reply #8 on: Jul 1^{st}, 2008, 1:27pm » 
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Thanks so much guys for pointing out some typos in the original question. I have revised accordingly. FYI, here is the revised one: Have A Great Day!


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wonderful
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Re: Interesting inequality
« Reply #9 on: Jul 1^{st}, 2008, 8:28pm » 
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Here is a more general version: Have A Great Day!


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pex
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Re: Interesting inequality
« Reply #10 on: Jul 3^{rd}, 2008, 2:32pm » 
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hidden:  By symmetry, we lose no generality in assuming 0 < x < y < z. Additionally, by homogeneity, we may set z = 1. What remains is a function of two variables x and y. I haven't explicitly checked it, but it looks like the function is everywhere increasing in x, so that the function approaches its infimum as x > 0. By continuity, we may set x = 0 for the moment to solve for y. The remaining function of one variable can be differentiated. After simplifying, we need to find the roots of a seventhdegree polynomial. Three of them are easy to locate (one is 1 and the others are the complex roots of x^{2}  x + 1); we are left with a fourthdegree polynomial. The roots of this polynomial can be found algebraically. The only one that lies between 0 and 1 is y = 3/4 + sqrt(5)/4  sqrt(6*sqrt(5)  2)/4. We calculate f(0, 3/4 + sqrt(5)/4  sqrt(6*sqrt(5)  2)/4, 1) = 5/2 + 5*sqrt(5)/2.  Thus, the greatest lower bound is k = 5/2 + 5*sqrt(5)/2, attained when x > 0, y = 3/4 + sqrt(5)/4  sqrt(6*sqrt(5)  2)/4, and z = 1. We observe that k is approximately equal to 8.0902 > 4.


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wonderful
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Re: Interesting inequality
« Reply #11 on: Jul 3^{rd}, 2008, 5:09pm » 
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Welldone Pex! You arrive at the correct conclusion. Can you find a simpler solution? More particularly, can you find a way to come up with a simpler maximization programming? Have A Great Day!


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wonderful
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Re: Interesting inequality
« Reply #12 on: Jul 4^{th}, 2008, 2:26pm » 
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Hi Pex, I looked at your solution and really like some the maximization techniques you used in the proof. Thanks for sharing. Have A Great Day! P.S. There are other solutions. If anyone find out, please feel free to post here.


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