some problem connected with prove homogeneous inequalities
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Pawel_Iks - 05 Oct 2008 21:31 GMT Let's consider real x,y,z,... from range (0,infinity). We are interested in true inequalities homogeneous of n-th order in variables x,y,z, ... The problem is to find all such inequalities.
1) for n=1, we have: A*(x+y)>=0, so the constant A is from S(A)={A: A>=0}
2) for n=2, we have: A*(x^2+y^2)+B*x*y>=0, it's easy to show that constants A,B have to satisfy inequality abs(B)>2*abs(A), so S(A,B)={(A,B): abs(A)>2*abs(B)}, and this set is a 2D cone with origin at (0,0) 3) for n=3, we have A*(x^3+y^3+z^3)+B*(x^2y+y^*x+...)+C*xyz>=0, I suppose that S(A,B,C) is also a cone with origin at (0,0,0), but I have no idea how to prove it 4) for n=4, we have: A(x^4+y^4+z^4+t^4)+B(xyz^2+...)+C(x^2y^2+...)+D(x^3y+...)+Exyzt>=0 I'm interesting if it's also a cone S(A,B,C,D,E) ...
and if it's a general rule. Are there some simply methods for finding such sets?
Robert Israel - 06 Oct 2008 00:25 GMT > Let's consider real x,y,z,... from range (0,infinity). We are > interested in true inequalities homogeneous of n-th order in variables [quoted text clipped - 15 lines] > A(x^4+y^4+z^4+t^4)+B(xyz^2+...)+C(x^2y^2+...)+D(x^3y+...)+Exyzt>=0 > I'm interesting if it's also a cone S(A,B,C,D,E) ... Of course: for any set S, the functions from S to [0, infty) form a cone. That is, if f_1 and f_2 are such functions and t_1, t_2 are in [0,infty), then t_1 f_1 + t_2 f_2 is such a function.
> and if it's a general rule. Are there some simply methods for finding > such sets?  Signature Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada
Robert Israel - 06 Oct 2008 03:06 GMT > > Let's consider real x,y,z,... from range (0,infinity). We are > > interested in true inequalities homogeneous of n-th order in variables > > x,y,z, ... The problem is to find all such inequalities. > > > > 1) for n=1, we have: > > A*(x+y)>=0, so the constant A is from S(A)={A: A>=0} Why not A*x + B*y?
> > 2) for n=2, we have: > > A*(x^2+y^2)+B*x*y>=0, it's easy to show that constants A,B > > have to satisfy inequality > > abs(B)>2*abs(A), so S(A,B)={(A,B): abs(A)>2*abs(B)}, and this > > set is a 2D cone with origin at (0,0) Are you assuming your function is not just homogeneous but also symmetric in the variables? How does the number of variables relate to n?
 Signature Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada
Pawel_Iks - 06 Oct 2008 14:36 GMT > Are you assuming your function is not just homogeneous but also symmetric in > the variables? Yes, of course. I've forgotten to mention it :(
> How does the number of variables relate to n? I don't think about it, however it could be computed in some combinatorial way, and I think that it isn't difficult. However, I'm interesting not in it, but how to compute the bounds of the set S (for n=2 we have the intersection of two half planes: y>=+-2x.
Pawel_Iks - 06 Oct 2008 14:43 GMT > > Are you assuming your function is not just homogeneous but also symmetric in > > the variables? > > Yes, of course. I've forgotten to mention it :( > > > How does the number of variables relate to n? i didn't note that you were asking about number of variables (I though that you are asking about constants A,B,...), for given number n, there are n variables, becouse I want to have term x_1*x_2*...*x_n.
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