11. 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.
11. First, prove $\frac{O A+O C}{O B+O D} \geqslant \frac{1}{\sqrt{2}}$. Squaring both sides of the inequality and eliminating the denominator, we get
$$
\begin{array}{l}
2\left(O A^{2}+O C^{2}+2 O A \cdot O C\right) \\
\geqslant O B^{2}+O D^{2}+2 O B \cdot O D .
\end{array}
$$

Using the well-known fact $O A^{2}+O C^{2}=O B^{2}+O C^{2}$, we can simplify it to $O B^{2}+O D^{2}+4 O A \cdot O C \geqslant 2 O B \cdot O D$. This inequality clearly holds, because $O B^{2}+O D^{2} \geqslant 2 O B \cdot O D$.

Second, if we take $O=A$, we see that the value of the fraction is $\frac{1}{\sqrt{2}}$. Therefore, the minimum value of the ratio is $\frac{1}{\sqrt{2}}$.