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inexorable
Full Member    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) IP Logged
Michael Dagg
Senior Riddler     Gender: Posts: 500 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
Uberpuzzler      I am no special. I am only passionately curious.

Gender: Posts: 1001 Re: prove the inequality   « Reply #2 on: Sep 12th, 2011, 2:04pm » Quote Modify

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

-- AI IP Logged

Self discovery comes when a man measures himself against an obstacle - Antoine de Saint Exupery
william wu       Gender: Posts: 1291 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
Uberpuzzler      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) ] IP Logged
william wu       Gender: Posts: 1291 Re: prove the inequality   « Reply #5 on: Mar 4th, 2012, 5:50pm » Quote Modify

Wow ... nice job. IP Logged

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Michael Dagg
Senior Riddler     Gender: Posts: 500 Re: prove the inequality   « Reply #6 on: Mar 9th, 2012, 5:34pm » Quote Modify

Very nice SWF. IP Logged

Regards,
Michael Dagg

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