2. Find all positive real triples $(x, y, z)$:
$$
\left\{\begin{array}{l}
x^{3}+2 y^{2}+\frac{1}{4 z}=1 \\
y^{3}+2 z^{2}+\frac{1}{4 x}=1 \\
z^{3}+2 x^{2}+\frac{1}{4 y}=1 .
\end{array}\right.
$$
(2016, Croatian Mathematical Competition)
$$
\begin{array}{l}
\frac{1}{4 x}\left(2 x^{2}+2 x-1\right)^{2}+\frac{1}{4 y}\left(2 y^{2}+2 y-1\right)^{2}+ \\
\frac{1}{4 z}\left(2 z^{2}+2 z-1\right)^{2}=0 .
\end{array}
$$

Since $x$, $y$, and $z$ are all positive, we have
$$
\begin{array}{l}
2 x^{2}+2 x-1=0, \\
2 y^{2}+2 y-1=0, \\
2 z^{2}+2 z-1=0 .
\end{array}
$$

Solving these, we get $x=y=z=\frac{-1+\sqrt{3}}{2}$.
$$
\begin{array}{l}
\text { Therefore, }(x, y, z) \\
=\left(\frac{-1+\sqrt{3}}{2}, \frac{-1+\sqrt{3}}{2}, \frac{-1+\sqrt{3}}{2}\right) \text {. }
\end{array}
$$