Greece — Selection Examination

a. Since $x \ge n^2$, we have
$$n\sqrt{x-n^2} \le \frac{x}{2} \Leftrightarrow 2n\sqrt{x-n^2} \le x \Leftrightarrow 4n^2(x-n^2) \le x^2 \Leftrightarrow (x-2n^2)^2 \ge 0,$$
which is valid. Equality holds if and only if $x = 2n^2$.

Alternatively, for every $x \ge n^2$, it is enough to prove that
$$n\sqrt{x-n^2} - \frac{x}{2} \le 0 \Leftrightarrow 2n\sqrt{x-n^2} - x \le 0 \Leftrightarrow \frac{(2n\sqrt{x-n^2} - x)(2n\sqrt{x-n^2} + x)}{2n\sqrt{x-n^2} + x} \le 0$$
$$\Leftrightarrow \frac{4n^2(x-n^2) - x^2}{2n\sqrt{x-n^2} + x} \le 0 \Leftrightarrow \frac{-(x-2n^2)^2}{2n\sqrt{x-n^2} + x} \le 0,$$
which is valid. Equality holds for $x = 2n^2$.

b. The given inequality can be written in the form
$$(2\sqrt{x-1}-x)+(4\sqrt{y-4}-y)+(6\sqrt{z-9}-z)=0, \quad (1)$$
for $x \ge 1$, $y \ge 4$ and $z \ge 9$.

Using (a) for $n = 1, 2, \dots, 3$, we get
$$2\sqrt{x-1}-x \le 0, \quad 4\sqrt{y-4}-y \le 0 \quad \text{and} \quad 6\sqrt{z-9}-z \le 0,$$
And therefore (1) is possible to be valid, only for
$$2\sqrt{x-1}-x=0, \quad 4\sqrt{y-4}-y=0 \quad \text{and} \quad 6\sqrt{z-9}-z=0 \Leftrightarrow x=2, y=8, z=18.$$