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Topic: x + y + z >= xyz implies x^2 + y^2 + z^2 >= Axyz
Replies: 20   Last Post: Nov 1, 2013 4:48 AM

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 dan.ms.chaos@gmail.com Posts: 409 Registered: 3/1/08
Re: x + y + z >= xyz implies x^2 + y^2 + z^2 >= Axyz
Posted: Oct 27, 2013 8:08 AM

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).

Date Subject Author
10/25/13 quasi
10/25/13 David C. Ullrich
10/25/13 quasi
10/26/13 Karl-Olav Nyberg
10/26/13 quasi
10/26/13 Peter Percival
10/25/13 David C. Ullrich
10/25/13 Don Coppersmith
10/25/13 Don Coppersmith
10/25/13 gnasher729
10/25/13 Peter Percival
10/27/13 gnasher729
10/27/13 dan.ms.chaos@gmail.com
10/27/13 Peter Percival
10/27/13 fom
10/30/13 quasi
10/30/13 Peter Percival
10/30/13 quasi
10/31/13 quasi
10/31/13 quasi
11/1/13 Phil Carmody