II. (25 points) Given that $G$ is any point inside $\triangle ABC$, and $BG$, $CG$ intersect $AC$, $AB$ at points $E$, $F$ respectively. Find the minimum value of $k$ such that the inequality $S_{\triangle BGF} \cdot S_{\triangle OCE} \leqslant k S_{\triangle ABC}^{2}$ always holds.
As shown in Figure 7, let
$$
\frac{A F}{A B}=x, \frac{A E}{A C}=y \text {. }
$$

If $0$ has real roots, then
$$
\Delta=(t-2)^{2}-8 t \geqslant 0
$$
$\Rightarrow t \geqslant 6+4 \sqrt{2}$ or $t \leqslant 6-4 \sqrt{2}$.
Thus, $t \leqslant 6-4 \sqrt{2}$.
Therefore, $t_{\max }=6-4 \sqrt{2}$, at this time $u=2 \sqrt{2}-2$.
Hence, when $u=2 \sqrt{2}-2, x=y$, i.e., $x=y=\sqrt{2}-1$,
$$
\begin{array}{l}
\left(\frac{S_{\triangle B F G} \cdot S_{\triangle C E G}}{S_{\triangle A B C}^{2}}\right)_{\max }=\frac{1}{4}(6-4 \sqrt{2})^{2} \\
=17-12 \sqrt{2} .
\end{array}
$$

Thus, $k \geqslant 17-12 \sqrt{2}, k_{\min }=17-12 \sqrt{2}$.