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        riddles >> putnam exam (pure math) >> Interesting inequality 
        
(Message started by: wonderful on Jun 30^th, 2008, 8:18pm)

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 z²+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:

Shouldn't the second term have z²+2xz in the numerator ?

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:

Undoubtedly. If so, I get f(x,y,z) > 3, but what do I know?

???

If we write f(x, y, z) = (y² + 2yz) / (y - z)² + (z² + 2xz) / (x - z)² + (x² + 2xy) / (x - y)²,
then isn't lim(m -> infinity) f(1, m, m²) = 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:

isn't lim(m -> infinity) f(1, m, m²) = 1?

That's what I was gonna say (well almost).

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:

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...

Title: Re: Interesting inequality
Post by towr on Jul 1^st, 2008, 7:19am

on 07/01/08 at 07:09:09, 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/m², 1/m, 1)
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, m²) -> 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 seventh-degree polynomial. Three of them are easy to locate (one is -1 and the others are the complex roots of x² - x + 1); we are left with a fourth-degree 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

Well-done 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.