4. (USA) Let $a, b, c$ be given positive constants.

Solve the system of equations
$$
\left\{\begin{array}{l}
x+y+z=a+b+c, \\
4 x y z-\left(a^{2} x+b^{2} y+c^{2} z\right)=a b c
\end{array}\right.
$$

for all positive real numbers $x, y, z$.
The second equation is equivalent to
$$
\begin{array}{l}
4=\frac{a^{2}}{y z}+\frac{b^{2}}{z x}+\frac{c^{2}}{x y}+\frac{a b c}{x y z} . \\
\text { Let } x_{1}=\frac{a}{\sqrt{y z}}, y_{1}=\frac{b}{\sqrt{z x}}, z_{1}= \\
4=x_{1}^{2}+y_{1}^{2}+z_{1}^{2}+x_{1} y_{1} z_{1} .
\end{array}
$$

where, $0<x_{1}<2,0<y_{1}<2,0<z_{1}<2$.
Considering the new equation as a quadratic equation in $z_{1}$, the expression $\left(4-x_{1}^{2}\right)\left(4-y_{i}^{2}\right)$ inspires us to set
$$
x_{1}=2 \sin u, 0<u<\frac{\pi}{2}, y_{1}=2 \sin v, 0<v<\frac{\pi}{2}
$$

Then, $4=4 \sin ^{2} u+4 \sin ^{2} v+z_{1}^{2}+4 \sin u \sin v z_{1}$.
Thus, $\left(z_{1}+2 \sin u \sin v\right)^{2}=4\left(1-\sin ^{2} u\right)(1-\sin ^{2} v)$.
That is, $\left|z_{1}+2 \sin u \sin v\right|=|2 \cos u \cos v|$. Therefore, we have
$$
z_{1}=2(\cos u \cos v-\sin u \sin v)=2 \cos (u+v) .
$$

Thus, $2 \sin u \sqrt{y z}=a, 2 \sin v \sqrt{z x}=b$,
$$
2(\cos u \cos v-\sin u \sin v) \sqrt{x y}=c .
$$

From $x+y+z=a+b+c$, we get
$$
\begin{array}{l}
(\sqrt{x} \cos v-\sqrt{y} \cos u)^{2}+(\sqrt{x} \sin v+\sqrt{y} \sin u-\sqrt{z})^{2}=0 .
\end{array}
$$

This leads to
$$
\begin{array}{l}
\sqrt{z}=\sqrt{x} \sin v+\sqrt{y} \sin u \\
=\sqrt{x} \frac{y_{1}}{2}+\sqrt{y} \frac{x_{1}}{2} .
\end{array}
$$

Therefore, $\sqrt{z}=\sqrt{x} \cdot \frac{b}{2 \sqrt{z x}}+\sqrt{y} \cdot \frac{a}{2 \sqrt{y z}}$, i.e., $z=\frac{1}{2}(a+b)$.
Similarly, we get $y=\frac{1}{2}(c+a), x=\frac{1}{2}(b+c)$.
Clearly, the triplet $(x, y, z)=\left(\frac{1}{2}(b+c), \frac{1}{2}(c+a), \frac{1}{2}(a+b)\right)$ satisfies the given system of equations, which is the unique solution.