Proposition 1 As shown in Figure 2, place three rectangles $X_{1}, X_{2}, X_{3}$ in an acute triangle $T$. Let $A(X)$ denote the area of rectangle $X$. Question: Does the maximum value of $\frac{A\left(X_{1}\right)+A\left(X_{2}\right)+A\left(X_{3}\right)}{A(T)}$ exist? If it exists, find the maximum value.
Solve as shown in Figure 2, let the side of rectangle $X_{1}$ that lies along the base of the triangle be denoted as the base of the triangle, with the base of the triangle being $a_{1}$, and the height of the triangle being $h$. The lengths of the sides of rectangles $X_{1}$, $X_{2}$, and $X_{3}$ parallel to the base of the triangle are denoted as $a_{2}$, $a_{3}$, and $a_{4}$, respectively, and the lengths of the sides perpendicular to the base of the triangle are denoted as $b_{1}$, $b_{2}$, and $b_{3}$, respectively. The distance from the vertex opposite the base to rectangle $X_{3}$ is denoted as $b_{4}$. Then,
$$
\begin{array}{c}
\frac{A\left(X_{1}\right)+A\left(X_{2}\right)+A\left(X_{3}\right)}{A(T)} \\
=\frac{a_{2} b_{1}+a_{3} b_{2}+a_{4} b_{3}}{\frac{h a_{1}}{2}},
\end{array}
$$

where, $h=b_{1}+b_{2}+b_{3}+b_{4}$.
From the similar triangles in Figure 2, we have
$$
\frac{a_{1}}{h}=\frac{a_{2}}{b_{2}+b_{3}+b_{4}}=\frac{a_{3}}{b_{3}+b_{4}}=\frac{a_{4}}{b_{4}} \text {. }
$$

Thus, $a_{2}=\frac{a_{1}\left(b_{2}+b_{3}+b_{4}\right)}{h}$,
$$
a_{3}=\frac{a_{1}\left(b_{3}+b_{4}\right)}{h}, a_{4}=\frac{a_{1} b_{4}}{h} .
$$

Substituting into equation (1), we get
$$
\begin{array}{l}
\frac{A\left(X_{1}\right)+A\left(X_{2}\right)+A\left(X_{3}\right)}{A(T)} \\
=\frac{\frac{a_{1}\left(b_{2}+b_{3}+b_{4}\right)}{h} b_{1}+\frac{a_{1}\left(b_{3}+b_{4}\right)}{h} b_{2}+\frac{a_{1} b_{4}}{h} b_{3}}{\frac{h a_{1}}{2}} \\
=\frac{2\left(b_{1} b_{2}+b_{1} b_{3}+b_{1} b_{4}+b_{2} b_{3}+b_{2} b_{4}+b_{3} b_{4}\right)}{h^{2}} \\
=\frac{\left(b_{1}+b_{2}+b_{3}+b_{4}\right)^{2}-\left(b_{1}^{2}+b_{2}^{2}+b_{3}^{2}+b_{4}^{2}\right)}{h^{2}} \\
\leqslant-\frac{h^{2}-\frac{1}{4}\left(b_{1}+b_{2}+b_{3}+b_{4}\right)^{2}}{h^{2}}=\frac{3}{4},
\end{array}
$$

with equality holding if and only if $b_{1}=b_{2}=b_{3}=b_{4}=\frac{h}{4}$. 
Therefore, the maximum value of $\frac{A\left(X_{1}\right)+A\left(X_{2}\right)+A\left(X_{3}\right)}{A(T)}$ is $\frac{3}{4}$.