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Topic: prove the inequality (Read 8933 times) 

inexorable
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prove the inequality
« on: Feb 3^{rd}, 2011, 1:39pm » 
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let x,y,z be positive numbers such that xyz=1 prove that (x^{5}+y^{5}+z^{5})^{2} >= 3(x^{7}+y^{7}+z^{7})


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Michael Dagg
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Re: prove the inequality
« Reply #1 on: Apr 6^{th}, 2011, 5:59pm » 
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I saw this a couple of months ago, I guess, and I thought sure that someone would have took some shots at it. The inequality does indeed hold. It is fairly involved to show, some algebraic artillery is helpful.

« Last Edit: Apr 6^{th}, 2011, 5:59pm by Michael Dagg » 
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TenaliRaman
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Re: prove the inequality
« Reply #2 on: Sep 12^{th}, 2011, 2:04pm » 
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I think just opening the square and applying Muirhead Inequality [1] does the job.  AI [1] https://nrich.maths.org/discus/messages/67613/Muirhead69859.pdf


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william wu
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Re: prove the inequality
« Reply #3 on: Jan 22^{nd}, 2012, 1:39pm » 
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Following TenaliRaman's idea: Let F(a,b,c) = x^a y^b z^c + x^a z^b y^c + y^a x^b z^c + y^a z^b x^c + z^a x^b y^c + z^a y^b x^c. Then LHS = (x^5+y^5+z^5)^2 = x^10 + y^10 + z^10 + 2 (x^5 y^5 + ... ) = (1/2) F(10,0,0) + F(5,5,0). and RHS = 3(x^7+y^7+z^7) = 3 xyz (x^7+y^7+z^7) = 3 (x^8 y z + y^8 x z + z^8 x y ) = (3/2) F(8,1,1) Comparing the LHS and the RHS, and multiplying both sides by 2, we want to show that F(10,0,0) + F(5,5,0) + F(5,5,0) >= F(8,1,1) + F(8,1,1) + F(8,1,1) Muirhead's Inequality shows that F(10,0,0) >= F(8,1,1), but it does not say anything about F(5,5,0) vs F(8,1,1) since neither sequence majorizes the other. So it seems that something more is needed here than Muirhead, unless there is some more algebraic preprocessing that can be done.

« Last Edit: Jan 22^{nd}, 2012, 1:40pm by william wu » 
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SWF
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Re: prove the inequality
« Reply #4 on: Mar 4^{th}, 2012, 10:02am » 
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This was a tough one! Without loss of generality, assume the values are named such that x >= y >= z > 0. To show that (x^{5}+y^{5}+z^{5})^{2}  3(x^{7}+y^{7}+z^{7}) >= 0 (with x*y*z=1), express it as the sum of a number of terms, all of which are positive or zero: xy(x  y)^{4}(x + y)^{2}(x^{2} + y^{2}) + yz(y  z)^{4}(y + z)^{2}(y^{2} + z^{2}) + xz(x  z)^{4}(x + z)^{2}(x^{2} + z^{2}) + (x  y)^{2}[ (2z^{4}  x^{4})^{2} + (2y^{4}  x^{4})^{2} ]/2 + (y  z)^{2}[ (2x^{4}  z^{4})^{2} + (2y^{4}  z^{4})^{2} ]/2 + (x  y)(y  z)[ (2z^{4}  x^{4})^{2} + 3(y^{8}  z^{8}) ]


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Michael Dagg
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Re: prove the inequality
« Reply #6 on: Mar 9^{th}, 2012, 5:34pm » 
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Very nice SWF.


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