2. Let $x, y, z$ be real numbers for which the following relations are true: $x=\sqrt{1-2 y z}, y=$ $\sqrt{1-2 z x}, z=\sqrt{1-2 x y}$. Calculate $x+y+z$.

Gazeta Matematică 11/2015.
## Solution.

Let $x, y, z \in \mathbb{R}$ with the properties given in the statement. Then $x, y, z \geq 0$, so $x+y+z \geq 0$. \mathbf{(2p)}.

From the hypothesis, $x^{2}+y^{2}+z^{2}=(1-2 y z)+(1-2 z x)+(1-2 x y)=3-2(x y+y z+z x)$. (1p) Then $(x+y+z)^{2}=3$, from which $x+y+z \in\{-\sqrt{3}, \sqrt{3}\}$. (3p)

From $x+y+z \geq 0$, it follows that $x+y+z=\sqrt{3}$. (1p)