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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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 quasi Posts: 12,067 Registered: 7/15/05
Re: x + y + z >= xyz implies x^2 + y^2 + z^2 >= Axyz
Posted: Oct 26, 2013 2:26 PM

konyberg wrote:
>quasi wrote:
>> dullrich wrote:
>> >quasi wrote:
>> >>
>> >>Here's a nice challenge problem which I adapted from a past
>> >>competition problem ...
>> >>
>> >>Problem:
>> >>
>> >>Find, with proof, the largest real number A such that
>> >>
>> >> x + y + z >= xyz
>> >>
>> >>implies
>> >>
>> >> x^2 + y^2 + z^2 >= Axyz

>> >
>> >I'm assuming you meant to restrict to x, y, z >= 0.

>>
>> No -- x,y,z are only required to be real.

>
>If x, y and z can be any real number then A can be any number.

No, not true.

>If x, y and z all equals 0 then A can be whatever it wants :)

Yes.

>So there must be some conditions to the problem.

No, only the conditions specified.

The problem is find the largest real number A such that
_all_ triples (x,y,z) which satisfy

x + y + z >= xyz

also satisfy

x^2 + y^2 + z^2 >= Axyz

quasi

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