G) 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 $\triangle A B C$.
First, find the minimum side length of an inscribed equilateral triangle within $\triangle ABC$.

As shown in Figure 1, 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 Law of Cosines, we have:
\[
\begin{array}{c}
D F^{2}=x^{2}+\left(1-\frac{x}{2}\right)^{2}-2 x\left(1-\frac{x}{2}\right) \cos 60^{\circ}= \\
\frac{7}{4} x^{2}-2 x+1 \\
D E^{2}=(1-x)^{2}+\left(\frac{\sqrt{3}}{2} x\right)^{2}=\frac{7}{4} x^{2}-2 x+1 \\
E F^{2}=\left(\sqrt{3}-\frac{\sqrt{3}}{2} x\right)^{2}+\left(1+\frac{x}{2}\right)^{2}- \\
2 \sqrt{3}\left(1-\frac{x}{2}\right)\left(1+\frac{x}{2}\right) \cos 30^{\circ}= \\
\frac{7}{4} x^{2}-2 x+1
\end{array}
\]

Thus, $D F = D E = E F$, indicating that $\triangle DEF$ 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 $\triangle DEF$ is denoted as $a$, then:
\[
\begin{aligned}
a^{2}= & \frac{7}{4} x^{2}-2 x+1=\frac{7}{4}\left(x^{2}-\frac{8}{7} x+\frac{4}{7}\right)= \\
& \frac{7}{4}\left(\left(x-\frac{4}{7}\right)^{2}-\frac{16}{49}+\frac{4}{7}\right)= \\
& \frac{7}{4}\left(x-\frac{4}{7}\right)^{2}+\frac{3}{7}
\end{aligned}
\]

It is clear that when $x = \frac{4}{7}$, the side length $a$ achieves its minimum value $\sqrt{\frac{3}{7}}$. The dashed triangle $\triangle PQR$ in Figure 1 represents the position where the minimum value is achieved.
Next, we prove that the longest side of any inscribed triangle is not less than $\sqrt{\frac{3}{7}}$. For convenience, we introduce a Cartesian coordinate system with $C$ as the origin and $CA$ as the positive $x$-axis (Figure 2). Let $\triangle XYZ$ be any inscribed triangle in $\triangle ABC$, where points $X$, $Y$, and $Z$ are on sides $BC$, $CA$, and $AB$ respectively. On $BM$, take points $R_1$ and $R_2$ such that $MR_1 = \frac{1}{3} MB$ and $MR_2 = \frac{3}{14} MB$. Thus, $AR_1 = \frac{2}{3} AB$ and $BR_2 = \frac{11}{14} BM$, leading to:
\[
\begin{array}{c}
y_{R_{1}}=\frac{2}{3}>\sqrt{\frac{3}{7}} \\
x_{R_{2}}=\frac{11}{14} \times \frac{\sqrt{3}}{2}>\frac{10}{13} \times \frac{\sqrt{3}}{2}>\sqrt{\frac{3}{7}}
\end{array}
\]
(1) If point $Z$ is on segment $BR_1$, then $y_Z \geq y_{R_1} > \sqrt{\frac{3}{7}}$; if point $Z$ is on segment $R_2A$, then $x_Z \geq x_{R_2} > \sqrt{\frac{3}{7}}$. Therefore, in this case, the longest side of $\triangle XYZ$ is greater than $\sqrt{\frac{3}{7}}$.
(2) If point $Z$ is within segment $R_1R_2$, then take point $Z$ as point $F$ and construct the equilateral $\triangle DEF$ as described at the beginning of the proof. It is easy to verify that $x_E < x_F$ and $y_D < y_F$. Thus, if point $X$ is on segment $DC$, then $ZX \geq ZD = FD$; if point $Y$ is on segment $CE$, then $ZY \geq ZE = FE$; if point $X$ is on $BD$ and point $Y$ is on $EA$, then by the Pythagorean theorem, $XY \geq DE$. Therefore, the longest side of $\triangle XYZ$ is not less than the side length of $\triangle DEF$, and thus not less than $\sqrt{\frac{3}{7}}$.
In summary, the minimum value of the longest side is $\sqrt{\frac{3}{7}}$.