Example 2 (17th All-Soviet Union Mathematical Olympiad, 9th Grade) One side of a rectangle is $1 \mathrm{~cm}$. It is known that it is divided into four smaller rectangles by two perpendicular lines, with the area of three of them being no less than $1 \mathrm{~cm}^{2}$, and the area of the fourth one being no less than $2 \mathrm{~cm}^{2}$. How long must the other side of the original rectangle be at least?
Let the side lengths of the small rectangle be as shown in Figure 4-1-1, then
$$
\left\{\begin{array}{l}
a+b=1, \\
a c \geqslant 1, \\
a d \geqslant 1, \\
b c \geqslant 1, \\
b d \geqslant 2 .
\end{array}\right.
$$

To find the minimum value of $c+d$, from the given conditions,
$$
c+d=(a+b)(c+d)=a c+b d+a d+b c \geqslant 1+2+2 \sqrt{a c \cdot b d} \geqslant 3+2 \sqrt{2} .
$$

The equality holds when $a=\sqrt{2}-1, b=2-\sqrt{2}, c=\sqrt{2}+1, d=2+\sqrt{2}$. 
Therefore, the other side is at least $3+2 \sqrt{2}$.