Example 11 In $\triangle A B C$, $\angle C=90^{\circ}, \angle A=$ $30^{\circ}, B C=1$, find the minimum value of the longest side of the inscribed triangle (a triangle with its three vertices on the three sides of $\triangle A B C$).
Solution: First, find the minimum side length of an inscribed equilateral triangle within $\triangle ABC$.

Take any point $D$ on side $BC$ and let $BD=x$. Then, take points $E$ and $F$ on sides $CA$ and $AB$ respectively, such that $CE=\frac{\sqrt{3}}{2} x$ and $BF=1-\frac{x}{2}$. By the cosine rule, we have
$$
\begin{aligned}
D F^{2}= & x^{2}+\left(1-\frac{x}{2}\right)^{2} \\
& -2 x\left(1-\frac{x}{2}\right) \cos 60^{\circ} \\
= & \frac{7}{4} x^{2}-2 x+1, \\
D E^{2}= & (1-x)^{2}+\left(\frac{\sqrt{3}}{2} x\right)^{2} \\
= & \frac{7}{4} x^{2}-2 x+1, \\
E F^{2}= & \left(\sqrt{3}-\frac{\sqrt{3}}{2} x\right)^{2}+\left(1+\frac{x}{2}\right)^{2} \\
& -2 \sqrt{3}\left(1-\frac{x}{2}\right)\left(1+\frac{x}{2}\right) \cos 30^{\circ}=\frac{7}{4} x^{2}-2 x+1 .
\end{aligned}
$$

Thus, $D F=D E=E F$, indicating that $\triangle D E F$ is an equilateral triangle. This shows that for any point $D$ on $BC$, an inscribed equilateral triangle can be constructed. Let the midpoints of $CA$ and $AB$ be $N$ and $M$ respectively, and the midpoint of $BM$ be $S$. As point $D$ moves from $B$ to $C$, point $E$ moves from $C$ to $N$, and point $F$ moves from $M$ to $S$. If the side length of $\triangle D E F$ is denoted as $a$, then
$$
\begin{aligned}
a^{2} & =\frac{7}{4} x^{2}-2 x+1 \\
& =\frac{7}{4}\left(x-\frac{4}{7}\right)^{2}+\frac{3}{7} .
\end{aligned}
$$

It is evident that when $x=\frac{4}{7}$, the side length $a$ achieves its minimum value of $\frac{1}{7} \sqrt{21}$. The dashed triangle $\triangle P Q R$ in the figure represents the position where the minimum value is achieved.
Next, we prove that the longest side of any inscribed triangle is not less than $\frac{1}{7} \sqrt{21}$.
For convenience, introduce a Cartesian coordinate system with $C$ as the origin and $CA$ as the positive $x$-axis. Let $\triangle X Y Z$ be any inscribed triangle in $\triangle A B C$, with points $X, Y, Z$ on sides $BC, CA, AB$ respectively. Take points $R_{1}$ and $R_{2}$ on $BM$ such that $M R_{1}=\frac{1}{3} M B$ and $M R_{2}=\frac{3}{14} M B$. Thus, $A R_{1}=\frac{2}{3} A B$ and $B R_{2}=\frac{11}{14} B M$, leading to
$$
\begin{array}{l}
y_{R_{1}}=\frac{2}{3}>\frac{1}{7} \sqrt{21}, \\
x_{R_{2}}=\frac{11}{14} \times \frac{\sqrt{3}}{2}>\frac{10}{13} \times \frac{\sqrt{3}}{2}>\frac{1}{7} \sqrt{21} .
\end{array}
$$
(1) If point $Z$ is on segment $B R_{1}$, then $y_{z} \geqslant y_{R_{1}}>\frac{1}{7} \sqrt{21}$; if point $Z$ is on segment $R_{2} A$, then $x_{x} \geqslant x_{R_{2}}>\frac{1}{7} \sqrt{21}$. Therefore, in this case, the longest side of $\triangle X Y Z$ is greater than $\frac{1}{7} \sqrt{21}$.
(2) Suppose point $Z$ is within segment $R_{1} R_{2}$. Then, take point $Z$ as point $F$ and construct the equilateral triangle $D E F$ as described at the beginning of the proof. It is not difficult to verify that $x_{E}<x_{F}$ and $y_{D}<y_{F}$. Therefore, if point $X$ is on segment $DC$, then $Z X \geqslant Z D=F D$; if point $Y$ is on segment $CE$, then $Z Y \geqslant Z E=F E$; if point $X$ is on $BD$ and point $Y$ is on $EA$, then by the Pythagorean theorem, $X Y \geqslant D E$. Thus, the longest side of $\triangle X Y Z$ is not less than the side length of $\triangle D E F$, and hence not less than $\frac{1}{7} \sqrt{21}$.
In conclusion, the minimum value of the longest side is $\frac{1}{7} \sqrt{21}$.