Height 185 Through the centroid $G$ of $\triangle A B C$, draw a line intersecting $A C$ and $A B$ at points $D$ and $E$ respectively, $B D$ and $C E$ intersect at point $P$. Let $k$ be the positive real number that for any $\triangle A B C, \triangle D E P$ can satisfy the inequality $S_{\triangle D E P} \leqslant k S_{\triangle A B C}$. Find the minimum value of $k$.
Solution: Let $\frac{A E}{A B}=x, \frac{A D}{A C}=y$. First, we prove $\frac{1}{x}+\frac{1}{y}=3$.
As shown in Figure 7, draw $B M / / D E$, and draw $C N / / D E$ intersecting line $A G$ at points $M$ and $N$ respectively, then
$$
\begin{array}{l}
\frac{A B}{A E}=\frac{A M}{A G}, \\
\frac{A C}{A D}=\frac{A N}{A G} .
\end{array}
$$

It is easy to prove $\triangle B M K \cong \triangle C N K$, so $M K=N K$. Then
$$
\begin{array}{l}
\frac{A B}{A E}+\frac{A C}{A D}=\frac{A M}{A G}+\frac{A N}{A G}=\frac{A M+A N}{A G} \\
=\frac{A K+M K+A K-N K}{A G}=\frac{2 A K}{A G} .
\end{array}
$$

Since $G$ is the centroid of $\triangle A B C$, then $A K=\frac{3}{2} A G$. Therefore,
$$
\frac{A B}{A E}+\frac{A C}{A D}=3,
$$

which means $\frac{1}{x}+\frac{1}{y}=3$.
In $\triangle A B D$, by Menelaus' theorem, we have
$\frac{B P}{P D} \cdot \frac{D C}{C A} \cdot \frac{A E}{E B}=1$.
Thus, $\frac{B P}{P D}=\frac{C A}{D C} \cdot \frac{E B}{A E}=\frac{1}{1-y} \cdot \frac{1-x}{x}=\frac{1-x}{x(1-y)}$, which means $\frac{B P}{B D}=\frac{1-x}{1-x y}$.
Similarly, $\frac{C P}{P E}=\frac{1-y}{y(1-x)}$.
And $\frac{S_{\triangle B B C}}{S_{\triangle B C D}}=\frac{B P}{B D}=\frac{1-x}{1-x y}, \frac{S_{\triangle B C D}}{S_{\triangle A B C}}=\frac{C D}{A C}=1-y$, then
$\frac{S_{\triangle B B C}}{S_{\triangle A B C}}=\frac{(1-x)(1-y)}{1-x y}$.
Also, $\frac{S_{\triangle D E P}}{S_{\triangle B P C}}=\frac{P D}{P B} \cdot \frac{P E}{P C}=\frac{x(1-y)}{1-x} \cdot \frac{y(1-x)}{1-y}=x y$,
thus, $\frac{S_{\triangle D B P}}{S_{\triangle B B C}}=\frac{(1-x)(1-y)}{1-x y} \cdot x y$
$=\frac{x y(1-x-y+x y)}{1-x y}$.
Since $x, y>0$, we have $x+y \geqslant 2 \sqrt{x y}$.
Therefore, $\frac{S_{\triangle D E P}}{S_{\triangle A B C}} \leqslant \frac{x y(1-2 \sqrt{x y}+x y)}{1-x y}=\frac{x y(1-\sqrt{x y})}{1+\sqrt{x y}}$.
From equation (1), we have $3=\frac{1}{x}+\frac{1}{y} \geqslant 2 \sqrt{\frac{1}{x y}}$,

which means $\sqrt{x y} \geqslant \frac{2}{3}$.
Also, $0 < x, y < 1$.
So, $f^{\prime}(u)<0$, which means $f(u)$ is a decreasing function.
Hence, $f(u) \leqslant f\left(\frac{2}{3}\right)=\frac{4}{45}$, which means $\frac{S_{\triangle D E P}}{S_{\triangle A B C}} \leqslant \frac{4}{45}$.

Therefore, $S_{\triangle D E P} \leqslant \frac{4}{45} S_{\triangle A B C}$.
Thus, $k_{\min }=\frac{4}{45}$.