Example 9 Solve the equation
\[
\begin{array}{l}
\frac{36}{\sqrt{x-2}}+\frac{4}{\sqrt{y-1}}+4 \sqrt{x-2}+\sqrt{y-1}-28 \\
=0 .
\end{array}
\]
(1988, Xuzhou City Junior High School Mathematics Competition)
Solution: The original equation can be reorganized as
$$
\begin{array}{l}
\left(4 \sqrt{x-2}+\frac{36}{\sqrt{x-2}}\right) \\
+\left(\sqrt{y-1}+\frac{4}{\sqrt{y-1}}\right)=28 .
\end{array}
$$

We know an important inequality: if $a \geqslant 0$, $b \geqslant 0$, then $a+b \geqslant 2 \sqrt{a b}$. Equality holds if and only if $a=b$. Applying this inequality, we have
$$
\begin{array}{l}
4 \sqrt{x-2}+\frac{36}{\sqrt{x-2}} \geqslant 24, \\
\sqrt{y-1}+\frac{4}{\sqrt{y-1}} \geqslant 4 .
\end{array}
$$

Thus, $4 \sqrt{x-2}+\frac{36}{\sqrt{x-2}}+\sqrt{y-1}+\frac{4}{\sqrt{y-1}}$
$\geqslant 28$.
Comparing this inequality with the original equation, we know that this inequality should take the equality. Therefore, we get
$$
\left\{\begin{array}{l}
4 \sqrt{x-2}=\frac{36}{\sqrt{x-2}}, \\
\sqrt{y-1}=\frac{4}{\sqrt{y-1}} .
\end{array}\right.
$$

Solving these equations, the real solutions of the original equation are $\left\{\begin{array}{l}x=11, \\ y=5 .\end{array}\right.$