439. Solve the system of equations:

$$
\left\{\begin{array}{l}
x+y+z=3 \\
x^{2}+y^{2}+z^{2}=3
\end{array}\right.
$$
$\triangle$ Using the left part of the first equation of the system, let's introduce vectors in space $\bar{u}(x ; y ; z)$ and $\bar{v}(1 ; 1 ; 1)$. We get:

$$
\bar{u} \cdot \bar{v}=x+y+z \leq|\bar{u}| \cdot|\bar{v}|=\sqrt{\left(x^{2}+y^{2}+z^{2}\right) 3}=\sqrt{3 \cdot 3}=3
$$

Since the left part of this inequality is also equal to 3, the inequality becomes an equality. Then $x=y=z$. Since $x+y+z=3$, the system of equations has a unique solution $(1 ; 1 ; 1)$.

Answer: $(1 ; 1 ; 1)$.