, 2008

Solution:
Kostas is right according to the following analysis:
If $x, y, z$ are the three positive real numbers Kostas thought about, then they satisfy the following equations:
$$\begin{gathered} x y + y z + z x = x + y + z \\ x y z = 1 \end{gathered}$$
Subtracting (1) from (2) by parts we obtain
$$\begin{gathered} x y z - (x y + y z + z x) = 1 - (x + y + z) \\ \Leftrightarrow x y z - x y - y z - z x + x + y + z - 1 = 0 \\ \Leftrightarrow x y(z-1) - x(z-1) - y(z-1) + (z-1) = 0 \\ \Leftrightarrow (z-1)(x y - x - y + 1) = 0 \\ (z-1)(x-1)(y-1) = 0 \\ \Leftrightarrow x=1 \text{ or } y=1 \text{ or } z=1 . \end{gathered}$$
For $x=1$, from (1) and (2) we have the equation $y z=1$, which has the solutions
$$(y, z) = \left(a, \frac{1}{a}\right), \ a>0$$
And therefore the solutions of the problem are the triples
$$(x, y, z) = \left(1, a, \frac{1}{a}\right), \ a>0$$
Similarly, considering $y=1$ or $z=1$ we get the solutions
$$(x, y, z) = \left(a, 1, \frac{1}{a}\right) \text{ or } (x, y, z) = \left(a, \frac{1}{a}, 1\right), \ a>0$$
Since for each $a>0$ we have
$$x + y + z = 1 + a + \frac{1}{a} \geq 1 + 2 = 3$$
and equality is valid only for $a=1$, we conclude that among the solutions of the problem, the triple $(x, y, z) = (1, 1, 1)$ is the one whose sum $x + y + z$ is minimal.