Question 127
Find all triplets of real positive numbers $x$, $y$ and $z$ such that
$$
\begin{cases}
\sqrt{2x - \frac{2}{y}} + \sqrt{2y - \frac{2}{z}} + \sqrt{2z - \frac{2}{x}} = \sqrt{3(x + y + z)}, \\
x^2 + y^2 + z^2 = 6.
\end{cases}
$$
$$
\begin{cases}
\sqrt{2x - \frac{2}{y}} + \sqrt{2y - \frac{2}{z}} + \sqrt{2z - \frac{2}{x}} = \sqrt{3(x + y + z)}, \\
x^2 + y^2 + z^2 = 6.
\end{cases}
$$