2. Since $1952=4 \cdot 488$, using the first two relations of the system, we get:

$$
\begin{aligned}
& \quad 25 x^{2}+20 y^{2}+13 z^{2} \leq 4 \cdot(3 x y+6 y z+4 z x) \Leftrightarrow 25 x^{2}+20 y^{2}+13 z^{2}-12 x y-24 y z-16 z x \leq 0 \Leftrightarrow \\
& \quad \Leftrightarrow\left(9 x^{2}-12 x y+4 y^{2}\right)+\left(16 y^{2}-24 y z+9 z^{2}\right)+\left(4 z^{2}-16 z x+16 x^{2}\right) \leq 0 \Leftrightarrow \\
& \Leftrightarrow(3 x-2 y)^{2}+(4 y-3 z)^{2}+(2 z-4 x)^{2} \leq 0 \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots
\end{aligned}
$$

From here it follows that $\frac{x}{2}=\frac{y}{3}=\frac{z}{4}=k \Rightarrow x=2 k, y=3 k, z=4 k$. The third relation of the system becomes: $2 k+3 k+4 k=18$, so $k=\frac{18}{9}=2$ and $x=2 \cdot 2=4, y=3 \cdot 2=6, z=4 \cdot 2=8$.

(Note: There is a minor error in the original text where it states $k=\frac{18}{2}=9$. It should be $k=\frac{18}{9}=2$.)
2. Determine $x, y, z \in \mathbb{R}$ that satisfy the relations: $\left\{\begin{array}{l}25 x^{2}+20 y^{2}+13 z^{2} \leq 1952 \\ 3 x y+6 y z+4 z x=488 \\ x+y+z=18\end{array}\right.$ Mihai Bodan, Cosmești, Teleorman