Example 6 Given real numbers $x, y, z$ satisfy
$$
\left\{\begin{array}{l}
x^{4}+y^{2}+4=5 y z, \\
y^{4}+z^{2}+4=5 z x, \\
z^{4}+x^{2}+4=5 x y .
\end{array}\right.
$$

Find the value of $x+y+z$.
$$
\begin{array}{l}
\left(x^{4}+y^{4}+z^{4}\right)+\left(x^{2}+y^{2}+z^{2}\right)-5(x y+y z+z x)+12 \\
=\left(x^{2}-2\right)^{2}+\left(y^{2}-2\right)^{2}+\left(z^{2}-2\right)^{2}+ \\
5\left[\left(x^{2}+y^{2}+z^{2}\right)-(x y+y z+z x)\right] \\
=\left(x^{2}-2\right)^{2}+\left(y^{2}-2\right)^{2}+\left(z^{2}-2\right)^{2}+ \\
\frac{5}{2}\left[(x-y)^{2}+(y-z)^{2}+(z-x)^{2}\right] \\
=0 \text {. } \\
\end{array}
$$

Using the property of non-negative numbers, we get
$$
x=y=z=\sqrt{2} \text { or }-\sqrt{2} \text {. }
$$

Therefore, $x+y+z=3 \sqrt{2}$ or $-3 \sqrt{2}$.