Romanian Mathematical Olympiad

The hypothesis can be written in the form $x(y-2) \le 2(z-2)$, $y(z-2) \le 2(x-2)$, $z(x-2) \le 2(y-2)$. If $x-2 < 0$, using the second relation we get $z-2 < 0$, then, according to the first, we obtain $y-2 < 0$, so $x$, $y$, $z \in (0, 2)$ and $xyz < 8$. We also have $x(2-y) \ge 2(2-z) > 0$, $y(2-z) \ge 2(2-x) > 0$, $z(2-x) \ge 2(2-y) > 0$. Now, multiplying the two inequalities and, after that, dividing by $(2-x)(2-y)(2-z) > 0$, we obtain $xyz \ge 8$, which is in contradiction with $xyz < 8$. Analogously, we obtain that, if $x > 2$, then $y > 2$ and $z > 2$, so $xyz > 8$. Multiplying these relations we get $xyz \le 8$ – contradiction. Therefore $x = 2$. Substituting in the initial conditions we obtain $y \le z \le 2 \le y$, so $x = y = z = 2$.

We have $ab + 4 - 2(a+b) = (a-2)(b-2)$. (1)
Suppose that one of the three numbers is less than $2$, for example $x < 2$. Then $yz + 4 \le 2(y+x) < 2y+4$, so $z < 2$. It follows from this that $xy + 4 \le 2(x+z) < 2x + 4$, so $y < 2$. From (1) we obtain $xy + 4 > 2(x+y)$ and the analogues, so $\sum xy + 12 > 4 \sum x$, which contradicts the hypothesis. Thus $x$, $y$, $z \ge 2$. From (1) we deduce that $xy + 4 \ge 2(x+y)$ and the analogues. By adding them we obtain $\sum xy + 12 \ge 4 \sum x$. Taking into account this relation and the hypothesis we deduce that all the previous inequalities must be equalities, so $x = y = z = 2$.