75th 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 $x=y=z=2$.

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Alternative solution.

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 $x=y=z=2$.