Greece — Selection Examination

First solution
We write the system in the form
$$xy + yz + zx = x + y + z \quad (1)$$
$$xyz = 1. \quad (2)$$
Subtracting the two equations by parts we find
$$\begin{aligned} & xyz - (xy + yz + zx) = 1 - (x + y + z) \\ \Leftrightarrow \quad & xyz - xy - yz - zx + x + y + z - 1 = 0 \\ \Leftrightarrow \quad & xy(z - 1) - x(z - 1) - y(z - 1) + (z - 1) = 0 \\ \Leftrightarrow \quad & (z - 1)(xy - x - y + 1) = 0 \\ \Leftrightarrow \quad & (z - 1)(x - 1)(y - 1) = 0 \\ \Leftrightarrow \quad & x = 1 \text{ or } y = 1 \text{ or } z = 1. \end{aligned}$$
For $x=1$, from (1) and (2) we have $yz=1$, which have the solutions
$$(y, z) = \left( a, \frac{1}{a} \right), \quad a > 0,$$
and so, the solutions of the system are
$$(x, y, z) = \left( 1, a, \frac{1}{a} \right), \quad a > 0.$$
Similarly, considering $y=1$ or $z=1$ we find the solutions
$$(x, y, z) = \left( a, 1, \frac{1}{a} \right) \text{ or } (x, y, z) = \left( a, \frac{1}{a}, 1 \right), \quad a > 0.$$
Since for each $a > 0$ we have
$$x + y + z = 1 + a + \frac{1}{a} \ge 1 + 2 = 3.$$
Equality holds for $a=1$, it follows that between the solutions of the system, the $(x, y, z) = (1, 1, 1)$ is that having the least possible sum $x+y+z$.

Second solution
Let $(x, y, z)$ is the solution of the system with the least possible sum. Then, from the inequality of arithmetic – geometric mean we have
$$\frac{x+y+z}{3} \ge \sqrt[3]{xyz} \Rightarrow x+y+z \ge 3,$$
while equality holds for $x = y = z$.
Hence the least possible value of the sum $x+y+z$, between the solution of the given system is 3 and it happens for $x = y = z$.
For $x = y = z$, from the equation $xyz = 1$, $x, y, z > 0$, it follows that $x = y = z = 1$, which satisfies also the equation $x + y + z = xy + yz + zx$.