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inexorable
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Posts: 211
 prove the inequality   « on: Feb 3rd, 2011, 1:39pm » Quote Modify

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 » Quote Modify

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 6th, 2011, 5:59pm by Michael Dagg » IP Logged

Regards,
Michael Dagg
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 » Quote Modify

I think just opening the square and applying Muirhead Inequality [1] does the job.

-- AI
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william wu

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 Re: prove the inequality   « Reply #3 on: Jan 22nd, 2012, 1:39pm » Quote Modify

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 22nd, 2012, 1:40pm by william wu » IP Logged

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SWF
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Posts: 879
 Re: prove the inequality   « Reply #4 on: Mar 4th, 2012, 10:02am » Quote Modify

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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william wu

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 Re: prove the inequality   « Reply #5 on: Mar 4th, 2012, 5:50pm » Quote Modify

Wow ... nice job.
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Michael Dagg
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 Re: prove the inequality   « Reply #6 on: Mar 9th, 2012, 5:34pm » Quote Modify

Very nice SWF.
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Regards,
Michael Dagg
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