Let $a, b, c$ be positive real numbers, determine all triples $\left\{\begin{array}{l}x+y \div z=a+b+c, \\ 4 x y z-\left(a^{2} x+b^{2} y+c^{2} z\right)=a b c .\end{array}\right.$ of $x, y, z$.
(36th IMO Shortlist)
It is not difficult to verify that $x=\frac{b+c}{2}, y=\frac{c+a}{2}, z=\frac{a+b}{2}$ is a solution. Below, we prove that this is the only solution.
$$
\text{Let } x=\frac{b+c}{2}+t_{1}, y=\frac{c+a}{2}+t_{2}, z=\frac{a+b}{2}+
$$
$t_{3}$, substituting into the original system of equations yields
$$
\left\{\begin{array}{l}
t_{1}+t_{2}+t_{3}=0, \\
2 t_{1} t_{2} t_{3}+(b+c) t_{2} t_{3}+(c+a) t_{3} t_{1}+(a+b) t_{1} t_{2}=0 .
\end{array}\right.
$$

If at least one of $t_{1}, t_{2}, t_{3}$ is 0, without loss of generality, assume $t_{1}=0$, then from (2) we get $(b+c) t_{2} t_{3}=0$.

Therefore, at least one of $t_{2}, t_{3}$ must be 0. By (1), we have $t_{1}=t_{2}=t_{3}=0$.

If $t_{1}, t_{2}, t_{3}$ are all non-zero, then at least two of $t_{1}, t_{2}, t_{3}$ must have the same sign. Without loss of generality, assume $t_{2}, t_{3}$ have the same sign. From (1) and (2), we get
$$
\begin{array}{l}
\left(2 t_{2}+c+a\right) t_{3}^{2}+2 t_{2} t_{3} a+\left(2 t_{3}+a+b\right) t_{2}^{2} \\
=0 .
\end{array}
$$

Since $2 t_{2}+c+a=2 y>0, 2 t_{3}+a+b=2 z>0, t_{2} t_{3}>0$, the left side of (3) is $>0$. This is a contradiction.
Therefore, the only solution that satisfies the conditions is $x=\frac{b+c}{2}$,
$$
y=\frac{c+a}{2}, z=\frac{a+b}{2} .
$$