4. A4 (USA) Let \( a, b \), and \( c \) be given positive real numbers. Determine all positive real numbers \( x, y \), and \( z \) such that 
\[ x+y+z=a+b+c \]
and 
\[ 4xyz - \left(a^2 x + b^2 y + c^2 z\right) = abc \]
4. The second equation is equivalent 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 $x_{1}=$ $\frac{a}{\sqrt{y z}}, y_{1}=\frac{b}{\sqrt{z x}}, z_{1}=\frac{c}{\sqrt{x y}}$. Then $x_{1}^{2}+y_{1}^{2}+z_{1}^{2}+x_{1} y_{1} z_{1}=4$, where $00$ is clearly impossible. On the other hand, if $u v w \leq 0$, then two of $u, v, w$ are nonnegative, say $u, v \geq 0$. Taking into account $w=-u-v$, the above equality reduces to $2\left[(a+c-2 v) u^{2}+(b+c-2 u) v^{2}+2 c u v\right]=0$, so $u=v=0$.  Third solution. The fact that we are given two equations and three variables suggests that this is essentially a problem on inequalities. Setting $f(x, y, z)=4 x y z-a^{2} x-b^{2} y-c^{2} z$, we should show that $\max f(x, y, z)=$ $a b c$, for $0<x, y, z, x+y+z=a+b+c$, and find when this value is attained. Thus we apply Lagrange multipliers to $F(x, y, z)=f(x, y, z)-$ $\lambda(x+y+z-a-b-c)$, and obtain that $f$ takes a maximum at $(x, y, z)$ such that $4 y z-a^{2}=4 z x-b^{2}=4 x y-c^{2}=\lambda$ and $x+y+z=a+b+c$. The only solution of this system is $(x, y, z)=\left(\frac{b+c}{2}, \frac{c+a}{2}, \frac{a+b}{2}\right)$.