Question 8 Let $a, b, c$ be given positive real numbers. Find all positive real numbers $x, y, z$ that satisfy the system of equations
$$
\left\{\begin{array}{l}
x+y+z=a+b+c, \\
4 x y z-\left(a^{2} x+b^{2} y+c^{2} z\right)=a b c .
\end{array}\right.
$$
(36th IMO Shortlist)
First, we prove the following lemma.

Lemma: The equation
$$
x^{2}+y^{2}+z^{2}+x y z=4
$$
has positive real solutions \(x, y, z\) if and only if there exists an acute triangle \(\triangle ABC\) such that \(x=2 \cos A, y=2 \cos B, z=2 \cos C\).

Proof of the lemma: First, by the identity (3), all triples \((2 \cos A, 2 \cos B, 2 \cos C)\) are solutions to equation (4).

Conversely, it is easy to see that \(0 < x, y, z < 2\), so there exist \(\angle A, \angle B \in \left(0, \frac{\pi}{2}\right)\) such that \(x = 2 \cos A, y = 2 \cos B\).
Solving equation (4) for \(z\) \((z \in (0,2))\), we get
$$
\begin{array}{l}
z = -2 \cos (A+B). \\
\text{Let } \angle C = \pi - \angle A - \angle B. \text{ Then} \\
(x, y, z) = (2 \cos A, 2 \cos B, 2 \cos C).
\end{array}
$$

Returning to the original problem.
Transform the second equation to \(\frac{a^{2}}{y z} + \frac{b^{2}}{z x} + \frac{c^{2}}{x y} + \frac{a b c}{x y z} = 4\).
Let \(u = \frac{a}{\sqrt{y z}}, v = \frac{b}{\sqrt{z x}}, w = \frac{c}{\sqrt{x y}}\). Then \(u^{2} + v^{2} + w^{2} + u v w = 4\).
According to the lemma, there exists an acute triangle \(\triangle ABC\) such that \(u = 2 \cos A, v = 2 \cos B, w = 2 \cos C\).
Thus, from the first equation, we have
$$
x + y + z = 2 \sqrt{x y} \cos C + 2 \sqrt{y z} \cos A + 2 \sqrt{z x} \cos B,
$$
which can be rewritten as
$$
\begin{array}{l}
(\sqrt{x})^{2} + (\sqrt{y})^{2} + (\sqrt{z})^{2} \\
= 2 \sqrt{x y} \cos C + 2 \sqrt{y z} \cos A + 2 \sqrt{z x} \cos B.
\end{array}
$$

This is the equality case of inequality (1).
Therefore, \(\frac{\sqrt{x}}{\sin A} = \frac{\sqrt{y}}{\sin B} = \frac{\sqrt{z}}{\sin C}\).
Using \(\cos A = \frac{a}{2 \sqrt{y z}}, \cos B = \frac{b}{2 \sqrt{z x}}, \cos C = \frac{c}{2 \sqrt{x y}}\)
and the cosine rule, we can solve for
$$
x = \frac{b + c}{2}, y = \frac{c + a}{2}, z = \frac{a + b}{2}.
$$

It is easy to verify that this triple satisfies the original system of equations.