Example 2 Given a square $A B C D$ in the plane. Find the minimum value of the ratio $\frac{O A+O C}{O B+O D}$, where $O$ is any point in the plane.
(1993, St. Petersburg City Mathematical Selection Exam)
Solution: As shown in Figure 2, it is easy to see that when $O$ is at point $A$, we have
$$
\begin{array}{l}
\frac{O A+O C}{O B+O D} \\
=\frac{A C}{A B+A D}=\frac{\sqrt{2}}{2} .
\end{array}
$$

When $O$ is inside the square, such as at the center of the square, the ratio is 1;

When $O$ is outside the square, such as on the extension of $D A$ and $O A = A D$, the obtained ratio is $\frac{1+\sqrt{5}}{2+\sqrt{2}}>\frac{\sqrt{2}}{2}$.
Thus, we conjecture that $\frac{\sqrt{2}}{2}$ might be the desired value.
The following proves that $\frac{O A+O C}{O B+O D} \geqslant \frac{\sqrt{2}}{2}$ (using proof by contradiction).
Assume $\frac{O A+O C}{O B+O D}<\frac{\sqrt{2}}{2}$, squaring and eliminating the denominator yields
$$
\begin{array}{l} 
2\left(O A^{2}+O C^{2}+2 O A \cdot O C\right) \\
<O B^{2}+O D^{2}+2 O B \cdot O D . \\
\because O A^{2}+O C^{2}=O B^{2}+O D^{2}, \\
\therefore O B^{2}+O D^{2}+4 O A \cdot O C<2 O B \cdot O D .
\end{array}
$$

However, $O B^{2}+O D^{2} \geqslant 2 O B \cdot O D$, which is a contradiction.
Therefore, $\frac{O A+O C}{O B+O D} \geqslant \frac{\sqrt{2}}{2}$,
which means the minimum value of $\frac{O A+O C}{O B+O D}$ is $\frac{\sqrt{2}}{2}$.