

Title: prove the inequality Post by inexorable on Feb 3^{rd}, 2011, 1:39pm 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}) 

Title: Re: prove the inequality Post by Michael Dagg on Apr 6^{th}, 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 12^{th}, 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/Muirhead69859.pdf 

Title: Re: prove the inequality Post by william wu on Jan 22^{nd}, 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 4^{th}, 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 (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}) ] 

Title: Re: prove the inequality Post by william wu on Mar 4^{th}, 2012, 5:50pm Wow ... nice job. 

Title: Re: prove the inequality Post by Michael Dagg on Mar 9^{th}, 2012, 5:34pm Very nice SWF. 

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