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Topic: prove the inequality (Read 8990 times) |
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inexorable
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prove the inequality
« on: Feb 3rd, 2011, 1:39pm » |
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let x,y,z be positive numbers such that xyz=1
prove that (x5+y5+z5)2 >= 3(x7+y7+z7)
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Michael Dagg
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Re: prove the inequality
« Reply #1 on: Apr 6th, 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.
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| « Last Edit: Apr 6th, 2011, 5:59pm by Michael Dagg » |
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Regards,
Michael Dagg
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TenaliRaman
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I am no special. I am only passionately curious.
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Re: prove the inequality
« Reply #2 on: Sep 12th, 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/Muirhead-69859.pdf
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Self discovery comes when a man measures himself against an obstacle - Antoine de Saint Exupery
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william wu
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Re: prove the inequality
« Reply #3 on: Jan 22nd, 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.
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| « Last Edit: Jan 22nd, 2012, 1:40pm by william wu » |
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SWF
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Posts: 879
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Re: prove the inequality
« Reply #4 on: Mar 4th, 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 (x5+y5+z5)2 - 3(x7+y7+z7) >= 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(x2 + y2) + yz(y - z)4(y + z)2(y2 + z2) + xz(x - z)4(x + z)2(x2 + z2)
+ (x - y)2[ (2z4 - x4)2 + (2y4 - x4)2 ]/2 + (y - z)2[ (2x4 - z4)2 + (2y4 - z4)2 ]/2
+ (x - y)(y - z)[ (2z4 - x4)2 + 3(y8 - z8) ]
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Michael Dagg
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Re: prove the inequality
« Reply #6 on: Mar 9th, 2012, 5:34pm » |
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Very nice SWF.
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Regards,
Michael Dagg
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