

Title: Interesting inequality Post by wonderful on Jun 30^{th}, 2008, 8:18pm http://www.imagehosting.com/out.php/i1816364_TTinequality.gif (http://www.imagehosting.com) Can you generalize the result? Have A Great Day! 

Title: Re: Interesting inequality Post by towr on Jul 1^{st}, 2008, 12:54am Shouldn't the second term have z^{2}+2xz in the numerator ? 

Title: Re: Interesting inequality Post by ThudanBlunder on Jul 1^{st}, 2008, 5:54am on 07/01/08 at 00:54:08, towr wrote:
Undoubtedly. If so, I get f(x,y,z) > 3, but what do I know? 

Title: Re: Interesting inequality Post by pex on Jul 1^{st}, 2008, 6:55am on 07/01/08 at 05:54:18, ThudanBlunder wrote:
??? 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? 

Title: Re: Interesting inequality Post by towr on Jul 1^{st}, 2008, 7:07am on 07/01/08 at 06:55:05, pex wrote:
Maybe we're supposed to assume x,y,z are integers? 

Title: Re: Interesting inequality Post by pex on Jul 1^{st}, 2008, 7:09am on 07/01/08 at 07:07:43, towr wrote:
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... 

Title: Re: Interesting inequality Post by towr on Jul 1^{st}, 2008, 7:19am on 07/01/08 at 07:09:09, pex wrote:
Quote:
Or rather, I picked two of them to be practically 0 (but of a different order) 

Title: Re: Interesting inequality Post by ThudanBlunder on Jul 1^{st}, 2008, 9:41am 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. http://www.imagehosting.com/out.php/i1815529_TTinequality.gif (http://www.imagehosting.com) 

Title: Re: Interesting inequality Post by wonderful on Jul 1^{st}, 2008, 1:27pm Thanks so much guys for pointing out some typos in the original question. I have revised accordingly. FYI, here is the revised one: http://www.imagehosting.com/out.php/i1816364_TTinequality.gif (http://www.imagehosting.com) Have A Great Day! 

Title: Re: Interesting inequality Post by wonderful on Jul 1^{st}, 2008, 8:28pm Here is a more general version: http://www.imagehosting.com/out.php/i1816781_TTinequality.gif (http://www.imagehosting.com) Have A Great Day! 

Title: Re: Interesting inequality Post by pex on Jul 3^{rd}, 2008, 2:32pm [hideb]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.[/hideb] Thus, the greatest lower bound is k = [hide]5/2 + 5*sqrt(5)/2[/hide], attained when [hide]x > 0, y = 3/4 + sqrt(5)/4  sqrt(6*sqrt(5)  2)/4, and z = 1[/hide]. We observe that k is approximately equal to [hide]8.0902[/hide] > 4. 

Title: Re: Interesting inequality Post by wonderful on Jul 3^{rd}, 2008, 5:09pm 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! 

Title: Re: Interesting inequality Post by wonderful on Jul 4^{th}, 2008, 2:26pm 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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