On Sunday, 27 October 2013 11:29:56 UTC+2, christian.bau wrote: > On Saturday, October 26, 2013 1:41:54 AM UTC+1, Peter Percival wrote: > > > > > A = sqrt 3. > > > > Proof?
I've managed to prove that the solution lies at the boundary:
(for each (x,y,z) we get a maximum A such that x^2 + y^2 + z^2 >= Axyz .) We are interested in the infimum of A over the set of (x,y,z) satisfying (x + y + z >= xyz ) . Assume the infimum A would occur for some (x,y,z) such that (x + y + z > xyz ) Then there is a k>1 such that (x + y + z = k^2 * xyz). (k*x,k*y,k*z) is of our set, therefore it must be so that (kx)^2 + (ky)^2 + (kz)^2 > A * (kx) *(ky) * (kz) .Taking out the k, we get that x^2 + y^2 + z^2 >= (k*A)xyz , k*A > A, contradicting the assumption that the maximum for these particular (x,y,z) (maximum >= k*A) is the infimum (A) .
By heuristic assumption that infimum occurs when x=y=z, we get A = sqrt(3).