## Task A-2.4.

Determine all triples of positive real numbers $(x, y, z)$ such that

$$
x^{3}+2 y^{2}+\frac{1}{4 z}=1, \quad y^{3}+2 z^{2}+\frac{1}{4 x}=1, \quad z^{3}+2 x^{2}+\frac{1}{4 y}=1
$$
## First Solution.

By adding the equations, we get

$$
x^{3}+2 x^{2}+\frac{1}{4 x}+y^{3}+2 y^{2}+\frac{1}{4 y}+z^{3}+2 z^{2}+\frac{1}{4 z}=3
$$

We will prove that $x^{3}+2 x^{2}+\frac{1}{4 x} \geqslant 1$. Multiplying by $4 x$ (which is positive), we see that this inequality is equivalent to $4 x^{4}+8 x^{3}+1 \geqslant 4 x$, or $\left(2 x^{2}+2 x-1\right)^{2} \geqslant 0$, which obviously holds. Therefore, for the left side in (*) to equal 3, the inequality we proved must actually be an equality, i.e., $2 x^{2}+2 x-1=0$. Solving this quadratic equation, we get $x=\frac{-1 \pm \sqrt{3}}{2}$. Since $a$ is positive, it follows that $x=\frac{-1+\sqrt{3}}{2}$. The same applies to the other variables, so $x=y=z=\frac{-1+\sqrt{3}}{2}$ is the only solution.