【Example 4】If $P, Q$ are on the perimeter of $\triangle A B C$, and the ratio of the area of the triangle cut off by $P Q$ to the area of $\triangle A B C$ is a constant $\lambda$, find the minimum length of $P Q$.

 untranslated part:
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。 

Note: The last sentence is a note about the translation request and is not part of the example text to be translated. It has been translated here for clarity.
Let $P$ be between $A$ and $B$, and $Q$ be between $A$ and $C$. Set $\angle A P Q=\alpha$, $\angle A Q P=\beta$, then
$$
\begin{aligned}
P Q^{2} & =\frac{2 \lambda \Delta \sin A}{\sin \alpha \sin \beta}=\frac{4 \lambda \Delta \sin A}{\cos (\alpha-\beta)+\cos A} \\
& \geqslant \frac{4 \lambda \Delta \sin A}{1+\cos A}=4 \lambda \Delta \tan \frac{A}{2} .
\end{aligned}
$$

Equality holds if and only if $\alpha=\beta$. At this point, $P Q^{2}$ reaches its minimum value $4 \lambda \Delta \tan \frac{A}{2}$. If $P, Q$ are on $B C$ and $A C$ respectively, similarly, the minimum value of $P Q^{2}$ is $4 \lambda \Delta \tan \frac{C}{2}$. If $P, Q$ are on $B C$ and $A B$ respectively, the minimum value of $P Q^{2}$ is $4 \lambda \Delta \tan ^{2} \frac{B}{2}$.

Assume $A \leqslant B \leqslant C$, then $\tan \frac{A}{2} \leqslant \tan \frac{B}{2} \leqslant \tan \frac{C}{2}$, so the smallest $P Q$ should be on the sides of the smallest angle, and $A P=A Q=\sqrt{\lambda b c}$.