Can xyz Exceed 1 Given x^2 + y^2 + z^2 + xyz = 4?

[anemone]

Here is this week's POTW:

-----

If $x,\,y$ and $z$ are non-negative reals such that $x^2+y^2+z^2+xyz=4$, prove that $xyz\le 1$.

-----

Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[anemone]

Congratulations to the following members for their correct solution::)

1. kaliprasad
2. lfdahl

Solution from kaliprasad:

Spoiler

Using the AM-GM inequality
$x^2+y^2+z^2 \ge 3\sqrt[3]{(xyz)^2}$
so $x^2+y^2+z^2+xyz\ge 3\sqrt[3]{(xyz)^2} + xyz $
hence $3\sqrt[3]{(xyz)^2} + xyz \le 4$
$3\sqrt[3]{a^2} + a \le 4$ where $a= xyz$ and from the given condition $a\ge 0$, we can say $3\sqrt[3]{a^2} + a$ is monotically increasing and at $a= 1$ we have $3\sqrt[3]{a^2} + a = 4$.

$\therefore 3\sqrt[3]{a^2} + a \le 4$ occurs at $a\le 1$ or $xyz \le 1$. (Q.E.D.)