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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 30, 2013 8:25 AM

quasi 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
>>
>> My solution is elementary (avoids Calculus), but if a method
>> based on Calculus yields an easy resolution, that would be worth
>> seeing as well.

>
> As a number of people have surmised, the maximum possible
> value of A is sqrt(3).
>
> Here's a proof ...
>
> Let b = sqrt(3).
>
> For x = y = z = b,
>
> x^2 + y^2 + z^2 = b(xyz)
>
> hence max(A) <= b.
>
> To show that max(A) = b, it suffices to show that for
> all real numbers x,y,z
>
> x + y + z >= xyz
>
> implies
>
> x^2 + y^2 + z^2 >= bxyz
>
> Suppose otherwise.
>
> Thus, assume x,y,z are real numbers such that
>
> x + y + z >= xyz
>
> x^2 + y^2 + z^2 < bxyz
>
> First
>
> 0 <= x^2 + y^2 + z^2
>
> => 0 < bxyz
>
> => xyz > 0
>
> => (x + y + z)^2 >= (xyz)^2
>
> Next
>
> (x-y)^2 + (y-z)^2 + (z-x)^2 >= 0
>
> => 2*(x^2 + y^2 + z^2) >= 2(xy + yz + zx)
>
> => 3(x^2 + y^2 + z^2) >= x^2 + y^2 + z^2 + 2(xy + yz + zx)
>
> => 3(x^2 + y^2 + z^2) >= (x + y + z)^2
>
> => 3(x^2 + y^2 + z^2) >= (xyz)^2
>
> => 3(bxyz) > (xyz)^2
>
> => xyz < 3b
>
> But
>
> x^2 + y^2 + z^2 >= 3(xyz)^(2/3) [by AM-GM]
>
> => bxyz > 3(xyz)^(2/3)
>
> => (bxyz)^3 > 27(xyz)^2
>
> => xyz > (27/b^3)
>
> => xyz > 3b
>
> contradiction, which completes the proof.
>
> quasi

That's neat. May I ask where the problem came from, and have you
considered similar with more than three numbers?

--
What a piece of work is a man. How noble in reason, how infinite
in faculties, in form and moving how express and admirable,
in action how like an angel, in apprehension how like a God
Shakespeare through the mouth of Hamlet

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