Example 9 Given that $a, b$ are positive numbers, and the parabolas $y=x^{2}+a x+2 b$ and $y=x^{2}+2 b x+a$ both have common points with the $x$-axis. Then the minimum value of $a^{2}+b^{2}$ is $\qquad$
$(2000$, National Junior High School Mathematics League)
Solution: From the given conditions, we have
$$
a^{2}-8 b \geqslant 0,4 b^{2}-4 a \geqslant 0 \text {. }
$$

Thus, $a^{4} \geqslant 64 b^{2} \geqslant 64 a$, which means $a \geqslant 4$.
Furthermore, $b^{2} \geqslant a \geqslant 4$. Therefore, $a^{2}+b^{2} \geqslant 20$.
Moreover, when $a=4, b=2$, the parabolas $y=x^{2}+a x+2 b$ and $y=x^{2}+2 b x+a$ both intersect the $x$-axis, so the minimum value of $a^{2}+b^{2}$ is 20.