For example, if $a$ and $b$ are positive numbers, and the parabolas $y=x^{2}+ax+2b$ and $y=x^{2}+2bx+a$ both intersect the $x$-axis. Then the minimum value of $a^{2}+b^{2}$ is $\qquad$
(2000, National Junior High School Mathematics League)
Solution: From the problem, we have
$$
\Delta_{1}=a^{2}-8 b \geqslant 0, \Delta_{2}=4 b^{2}-4 a \geqslant 0 \text {. }
$$

Thus, $a^{2} \geqslant 8 b$ and $b^{2} \geqslant a$.
Since $a$ and $b$ are both positive numbers,
$$
\therefore a^{4} \geqslant 64 b^{2} \geqslant 64 a \text {, i.e., } a \geqslant 4 \text {. }
$$

Similarly, we have $b^{2} \geqslant a \geqslant 4, b \geqslant 2$.
Therefore, $a_{\min }=4, b_{\text {min }}=2$.
$$
\text { Hence }\left(a^{2}+b^{2}\right)_{\min }=4^{2}+2^{2}=20 \text {. }
$$