Example 3 Real numbers $x, y, z$ satisfy
$$
x+y+z=4(\sqrt{x-5}+\sqrt{y-4}+\sqrt{z-3}) \text {. }
$$

Then $x=$ $\qquad$ ,$y=$ $\qquad$ ,$z=$ $\qquad$ .
(10th Five Sheep Cup Junior High School Mathematics Competition)
Solution: From the original equation, we have
$$
\begin{array}{l}
(\sqrt{x-5}-2)^{2}+(\sqrt{y-4}-2)^{2}+(\sqrt{z-3}-2)^{2}=0 . \\
\text { Also, } (a-b)^{2} \geqslant 0, \text { so } \\
\sqrt{x-5}-2=0, \sqrt{y-4}-2=0, \\
\sqrt{z-3}-2=0 .
\end{array}
$$
$$
\text { Therefore, } x=9, y=8, z=7 \text {. }
$$