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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 ]
 quasi Posts: 12,067 Registered: 7/15/05
Re: x + y + z >= xyz implies x^2 + y^2 + z^2 >= Axyz
Posted: Oct 30, 2013 4:12 PM

Peter Percival wrote:
>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.

>
>That's neat. May I ask where the problem came from,

I don't remember the source.

It was definitely a competition problem -- I think a Junior
level contest (but a strong one).

What I remember is that the problem was to prove:

If x + y + z >= xyz then x^2 + y^2 + z^2 >= xyz

From my solution, it became clear that the conclusion could be
strengthened from

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

to

x^2 + y^2 + z^2 >= sqrt(3)*(xyz)

>and have you considered similar with more than three numbers?

My solution generalizes directly to prove the following:

For n > 1, the largest real number A such that

x_1 + ... + x_n >= (x_1) * ... * (x_n)

implies

(x_1)^2 + ... + (x_n)^2 >= A*(x_1) * ... * (x_n)

is

A = n^(1/(n-1))

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