Example 3 Find the real solutions of the equation $x \sqrt{y-1}+y \sqrt{x-1}=x y$.

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Given $y \geqslant 1$, we think of
$$
y=1+(y-1) \geqslant 2 \sqrt{y-1} \text {, }
$$

with equality holding if and only if $y=2$.
Thus, $x y \geqslant 2 x \sqrt{y-1}$.
Similarly, $x y \geqslant 2 y \sqrt{x-1}$.
Therefore, $2 x y \geqslant 2 x \sqrt{y-1}+2 y \sqrt{x-1}$
$$
\Rightarrow x y \geqslant x \sqrt{y-1}+y \sqrt{x-1} \text {, }
$$

with equality holding if and only if $x=2$ and $y=2$.
【Note】From the condition, we can also think of
$$
x=\sec ^{2} \alpha, y=\sec ^{2} \beta\left(0 \leqslant \alpha, \beta \leqslant \frac{\pi}{2}\right),
$$

or set $a=\sqrt{x-1}, b=\sqrt{y-1}$.
Thus, we have
Alternative Solution 1 Given $x, y \geqslant 1$, let
$$
x=\sec ^{2} \alpha, y=\sec ^{2} \beta\left(0 \leqslant \alpha, \beta \leqslant \frac{\pi}{2}\right) \text {. }
$$

Then $\sec ^{2} \alpha \cdot \tan \beta+\sec ^{2} \beta \cdot \tan \alpha=\sec ^{2} \alpha \cdot \sec ^{2}$
$$
\begin{array}{l}
\Rightarrow \sin 2 \alpha+\sin 2 \beta=2 \\
\Rightarrow \sin 2 \alpha=1, \sin 2 \beta=1 \\
\Rightarrow x=y=2 .
\end{array}
$$

Alternative Solution 2 Set $a=\sqrt{x-1}, b=\sqrt{y-1}$. Then
$$
\begin{array}{l}
\left(a^{2}+1\right) b+\left(b^{2}+1\right) a=\left(a^{2}+1\right)\left(b^{2}+1\right) \\
(a-1)^{2}+(b-1)^{2}+a^{2}(b-1)^{2}+b^{2}(a-1)^{2}=
\end{array}
$$

Solving, we get $a=b=1$.
Thus, $x=y=2$.