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wonderful
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Interesting inequality
« on: Jun 30th, 2008, 8:18pm » |
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Can you generalize the result?
Have A Great Day!
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| « Last Edit: Jul 1st, 2008, 1:25pm by wonderful » |
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towr
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Re: Interesting inequality
« Reply #1 on: Jul 1st, 2008, 12:54am » |
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Shouldn't the second term have z2+2xz in the numerator ?
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ThudnBlunder
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Re: Interesting inequality
« Reply #2 on: Jul 1st, 2008, 5:54am » |
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on Jul 1st, 2008, 12:54am, towr wrote:| Shouldn't the second term have z2+2xz in the numerator ? |
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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 1st, 2008, 6:55am » |
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on Jul 1st, 2008, 5:54am, ThudanBlunder wrote:
Undoubtedly. If so, I get f(x,y,z) > 3, but what do I know?
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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?
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towr
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Re: Interesting inequality
« Reply #4 on: Jul 1st, 2008, 7:07am » |
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on Jul 1st, 2008, 6:55am, pex wrote:| isn't lim(m -> infinity) f(1, m, m2) = 1? |
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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 1st, 2008, 7:09am » |
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on Jul 1st, 2008, 7:07am, towr wrote:| Maybe we're supposed to assume x,y,z are integers? |
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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...
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| « Last Edit: Jul 1st, 2008, 7:10am by pex » |
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towr
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Re: Interesting inequality
« Reply #6 on: Jul 1st, 2008, 7:19am » |
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on Jul 1st, 2008, 7:09am, pex wrote:| Wouldn't the same counterexample still work? |
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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... |
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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)
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ThudnBlunder
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Re: Interesting inequality
« Reply #7 on: Jul 1st, 2008, 9:41am » |
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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.
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wonderful
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Re: Interesting inequality
« Reply #8 on: Jul 1st, 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 1st, 2008, 8:28pm » |
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Here is a more general version:
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pex
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Re: Interesting inequality
« Reply #10 on: Jul 3rd, 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 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. |
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 3rd, 2008, 5:09pm » |
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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!
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wonderful
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Re: Interesting inequality
« Reply #12 on: Jul 4th, 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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