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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

 Messages: [ Previous | Next ]
 Peter Percival Posts: 2,623 Registered: 10/25/10
Re: x + y + z >= xyz implies x^2 + y^2 + z^2 >= Axyz
Posted: Oct 26, 2013 4:09 PM

konyberg wrote:
> On Friday, October 25, 2013 7:50:36 PM UTC+2, 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.
>>
>>
>>
>> quasi

>
> If x, y and z can be any real number then A can be any number. If x, y and z all equals 0 then A can be whatever it wants :) So there must be som conditions to the problem.

At the point I've put ************************** above imagine universal
quantifiers binding x, y and z. Yes, if x = y = z = 0 then A can be
anything, but an A is required that works for all x, y, z.

--
The world will little note, nor long remember what we say here
Lincoln at Gettysburg

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