(British MO 1998)

Solve for $x, y, z \in \mathbb{R}^{+}$ real positive numbers the following system of equations:

$$
\begin{aligned}
& x y + y z + z x = 12 \\
& x y z = 2 + x + y + z
\end{aligned}
$$
This exercise is a disguised inequality problem. We notice that $(2,2,2)$ is a solution, so we would like to show that we are in the case of equality of an inequality.

Indeed, by AM-GM we have

$$
12 \geq x y+y z+z x \geq 3 \sqrt[3]{x^{2} y^{2} z^{2}}
$$

as well as

$$
x y z=2+x+y+z \geq 2+3 \sqrt[3]{x y z}
$$

We see that if we let $X=\sqrt[3]{x y z}$, both $X^{2} \leq 4$ and $X^{3}-3 X-2 \geq 0$, which means $X \leq 2$ and $(X-2)(X+1)^{2} \geq 0$, which is equivalent to saying that $X=2$. Thus, all the inequalities are equalities, and we quickly obtain $x=y=z=2$, the only solution.

The following exercises have not been covered in class.