## Task A-2.2.

Determine all real solutions of the system

$$
\begin{aligned}
& \left(1+4 x^{2}\right) y=4 z^{2} \\
& \left(1+4 y^{2}\right) z=4 x^{2} \\
& \left(1+4 z^{2}\right) x=4 y^{2}
\end{aligned}
$$
## Solution.

If any of $x, y, z$ is equal to zero, then the others are as well, and this is one solution.

Otherwise, it must be $x>0, y>0, z>0$.

By multiplying the given equalities, we get

$$
x y z\left(1+4 x^{2}\right)\left(1+4 y^{2}\right)\left(1+4 z^{2}\right)=64 x^{2} y^{2} z^{2}
$$

or, due to $x y z \neq 0$,

$$
\left(1+4 x^{2}\right)\left(1+4 y^{2}\right)\left(1+4 z^{2}\right)=64 x y z
$$

Since for positive $x, y, z$ we have $1+4 x^{2} \leqslant 4 x, 1+4 y^{2} \leqslant 4 y, 1+4 z^{2} \leqslant 4 z$, it follows that

$$
\left(1+4 x^{2}\right)\left(1+4 y^{2}\right)\left(1+4 z^{2}\right) \leqslant 64 x y z
$$

with equality holding only for $x=y=z=\frac{1}{2}$.

Thus, the only solutions to the given system are $(x, y, z)=(0,0,0)$ and $(x, y, z)=\left(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}\right)$.

## Second Solution.

Since the expressions in parentheses and on the right-hand sides of all equalities are non-negative, it must be $x \geqslant 0, y \geqslant 0, z \geqslant 0$.

We have $1+4 x^{2} \geqslant 4 x$, so

$$
4 z^{2}=\left(1+4 x^{2}\right) y \geqslant 4 x y
$$

thus $z^{2} \geqslant x y$.

Similarly, we get $x^{2} \geqslant y z$ and $y^{2} \geqslant z x$.

By multiplying these inequalities, we obtain $x^{2} y^{2} z^{2} \geqslant x^{2} y^{2} z^{2}$, and since equality holds here, we conclude that equality holds in each of the previous inequalities as well.

This means $x y=z^{2}, y z=x^{2}, z x=y^{2}$.

If one of the numbers is zero, then the others are as well.

Otherwise, dividing the first two equalities gives $\frac{x}{z}=\frac{z^{2}}{x^{2}}$, or $z=x$. Similarly, $x=y$.

Since $x=y=z$, we have $\left(1+4 x^{2}\right) x=4 x^{2}$.

Given $x \neq 0$, we have $4 x^{2}-4 x+1=0$. The solution to this equation is $x=\frac{1}{2}$.

The only solutions to the given system are $(0,0,0)$ and $\left(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}\right)$.