From: israel@math.ubc.ca (Robert Israel) Subject: Re: inequality Date: 23 Aug 2000 09:09:21 GMT Newsgroups: sci.math Summary: [missing] In article <8nt6lo$a4l$1@gannett.math.niu.edu>, Dave Rusin wrote: >(3) Am I correct, that (u^2+v^2) - 3*(u*v)^2 + (u*v)^4 is not a sum > of squares of polynomials? Yes. If u^2+v^2-3(uv)^2+(uv)^4 was a sum of squares of polynomials P_j, each of these P_j would have degree at most 2 in u and degree at most 2 in v, and the only terms of degree 2 in u (or similarly in v) could be a constant multiple of u^2 v^2. There is no constant term because P_j(0,0) = 0. This leaves P_j of the form a u^2 v^2 + b u v + c u + d v. Plug this into the equations P_j(1,1)=P_j(-1,1)=P_j(1,-1)=P_j(-1,-1)=0 and you get a=b=c=d=0. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ============================================================================== From: gwang Subject: Some Obscure Inequalities Date: Tue, 29 Aug 2000 21:39:22 -0400 Newsgroups: sci.math Summary: [missing] Hello, Does anyone out there know of a proof of these two inequalities: Polya's Inequality: (a - b)/[ln(a)-ln(b)] < (2sqrt(ab) + (a/2 + b/2))/3 where a and b are positive reals. I'm not sure about this, but I think a > b > 0. Carlson's Inequality: sqrt[(ab + bc + ca)/3] <= [cuberoot((a + b)(b + c)(c + a))]/2 where a, b, c are non-negative reals. Also while on inequalities, one of my high school friends went to a math camp this summer and said that he learned something called Muirhead's Theorem which was supposed to be very powerful for proving symmetric inequalities. However, his explanation of this theorem was less than clear. I have tried to find this theorem on the net and in a few math books, but is still unclear as to exactly what this theorem is. Any info on this theorem would be greatly appreciated. TIA, Xi Wang ============================================================================== From: taga@news.uni-rostock.de (Dr. Michael Ulm) Subject: Re: Some Obscure Inequalities Date: 30 Aug 2000 08:58:22 +0200 Newsgroups: sci.math On Tue, 29 Aug 2000 21:39:22 -0400, gwang wrote: >Hello, > >Does anyone out there know of a proof of these two inequalities: >Polya's Inequality: (a - b)/[ln(a)-ln(b)] < (2sqrt(ab) + (a/2 + b/2))/3 >where a and b are positive reals. I'm not sure about this, but I think >a > b > 0. >Carlson's Inequality: >sqrt[(ab + bc + ca)/3] <= [cuberoot((a + b)(b + c)(c + a))]/2 where a, >b, c are non-negative reals. Polyas Inequality is easy to prove (though a bit technical). The condition a > b > 0 is not necessary since the inequality is symmetric. But due to this symmetry, we may assume a > b > 0. Divide the inequality by b and set s^2 = a/b. Then all we have to prove is the simple one-dimensional inequality (s^2 - 1)/ log(s) < (1 + 4 s + s^2)/3 for s > 1. A little rearangement of terms gives log(s) - 3 (s^2 - 1)/(1 + 4 s + s^2 ) > 0 for s > 1. The left hand side is a function in s which is equal to 0 if s = 1, and its derivative is 1/s - 12 (1 + s + s^2)/(1 + 4 s + s^2)^2 = (s - 1)^4 /( s (1 + 4 s + s^2)^2) > 0. For Carlsons Inequality I found the following proof. Set s1 = a + b + c s2 = ab + bc + ac s3 = abc then the following inequalities are the well known symmetric mean inequalities s1 / 3 >= (s2 / 3)^(1/2) >= s3^(1/3). Carlson's Inequality is (s2 / 3)^(1/2) <= (s1 s2 - s3)^(1/3)/2. Use the symmetric mean inequalities to obtain s1 s2 - s3 >= 3 (s2 / 3)^(1/2) s2 - (s2 / 3)^(3/2) = 8 (s2 / 3)^(3/2). Carlsons Inequality immediately follows. HTH, Michael. -- &&&&&&&&&&&&&&&&#@#&&&&&&&&&&&&&&&& Dr. Michael Ulm FB Mathematik, Universitaet Rostock taga@sun15.math.uni-rostock.de