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        riddles >> putnam exam (pure math) >> prove the inequality
        
(Message started by: inexorable on Feb 3rd, 2011, 1:39pm)
Title: prove the inequality
Post by inexorable on Feb 3rd, 2011, 1:39pm
let x,y,z be positive numbers such that xyz=1
prove that (x5+y5+z5)2 >= 3(x7+y7+z7)
Title: Re: prove the inequality
Post by Michael Dagg on Apr 6th, 2011, 5:59pm
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.
Title: Re: prove the inequality
Post by TenaliRaman on Sep 12th, 2011, 2:04pm
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
Title: Re: prove the inequality
Post by william wu on Jan 22nd, 2012, 1:39pm
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.
Title: Re: prove the inequality
Post by SWF on Mar 4th, 2012, 10:02am
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) ]
Title: Re: prove the inequality
Post by william wu on Mar 4th, 2012, 5:50pm
Wow ... nice job.
Title: Re: prove the inequality
Post by Michael Dagg on Mar 9th, 2012, 5:34pm
Very nice SWF.


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