Author 
Topic: prove the inequality (Read 8926 times) 

inexorable
Full Member
Posts: 211


prove the inequality
« on: Feb 3^{rd}, 2011, 1:39pm » 
Quote Modify

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})


IP Logged 



Michael Dagg
Senior Riddler
Gender:
Posts: 500


Re: prove the inequality
« Reply #1 on: Apr 6^{th}, 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 6^{th}, 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 12^{th}, 2011, 2:04pm » 
Quote Modify

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


IP Logged 
Self discovery comes when a man measures himself against an obstacle  Antoine de Saint Exupery



william wu
wu::riddles Administrator
Gender:
Posts: 1291


Re: prove the inequality
« Reply #3 on: Jan 22^{nd}, 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 22^{nd}, 2012, 1:40pm by william wu » 
IP Logged 
[ wu ] : http://wuriddles.com / http://forums.wuriddles.com



SWF
Uberpuzzler
Posts: 879


Re: prove the inequality
« Reply #4 on: Mar 4^{th}, 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 (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}) ]


IP Logged 



Michael Dagg
Senior Riddler
Gender:
Posts: 500


Re: prove the inequality
« Reply #6 on: Mar 9^{th}, 2012, 5:34pm » 
Quote Modify

Very nice SWF.


IP Logged 
Regards, Michael Dagg



