II. (20 points) As shown in Figure 1, it is known that $M$ is a moving point on the line where side $DC$ of the square $ABCD$ lies. Find the maximum value of $\frac{MA}{MB}$.

保留源文本的换行和格式，翻译结果如下：

```
II. (20 points) As shown in Figure 1, it is known that $M$ is a moving point on the line where side $DC$ of the square $ABCD$ lies. Find the maximum value of $\frac{MA}{MB}$.
```
Let's assume the side length of the square is 2, and establish a Cartesian coordinate system with $DC$ as the $x$-axis and the midpoint of $DC$ as the origin $O$.
Then $A(-1,2)$ and $B(1,2)$.
Let $M(x, 0)$.
When $M A$ reaches its maximum value, it is clear that $x>0$.
$$
\begin{array}{l}
\text { Also, }\left(\frac{M A}{M B}\right)^{2}=\frac{(x+1)^{2}+4}{(x-1)^{2}+4}=1+\frac{4 x}{x^{2}-2 x+5} \\
=1+\frac{4}{x+\frac{5}{x}-2} \leqslant 1+\frac{4}{2 \sqrt{5}-2}=\frac{3+\sqrt{5}}{2}, \\
\text { thus, } \frac{M A}{M B} \leqslant \sqrt{\frac{3+\sqrt{5}}{2}}=\frac{\sqrt{5}+1}{2} .
\end{array}
$$

When $x=\sqrt{5}$, the equality holds.
Therefore, the maximum value of $\frac{M A}{M B}$ is $\frac{\sqrt{5}+1}{2}$.