Example 6 Given positive numbers $a_{1}, a_{2}, \cdots, a_{n} ; b_{1}, b_{2}$, $\cdots, b_{n}$ satisfying
$$
a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}=b_{1}^{2}+b_{2}^{2}+\cdots+b_{n}^{2}=1 .
$$

Find the maximum value of $F=\min \left\{\frac{a_{1}}{b_{1}}, \frac{a_{2}}{b_{2}}, \cdots, \frac{a_{n}}{b_{n}}\right\}$.
(1979, Guangdong Province High School Mathematics Competition)
Solution: It is easy to see that when all the letters are equal, the value of $F$ is 1.

Below we prove that for any positive numbers $a_{1}, a_{2}, \cdots, a_{n}$; $b_{1}, b_{2}, \cdots, b_{n}$, we have $F \leqslant 1$.
If not, then $F>1$.
$$
\text { Hence } \frac{a_{1}}{b_{1}}>1, \frac{a_{2}}{b_{2}}>1, \cdots, \frac{a_{n}}{b_{n}}>1 \text {, }
$$

which implies $a_{1}^{2}>b_{1}^{2}, a_{2}^{2}>b_{2}^{2}, \cdots, a_{n}^{2}>b_{n}^{2}$.
Thus, $a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}>b_{1}^{2}+b_{2}^{2}+\cdots+b_{n}^{2}$, contradicting the given condition.
Therefore, the maximum value of $F$ is 1.