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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 30, 2013 5:50 AM

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

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