Example 1 Given that $a, b$ are positive real numbers, the equations
$$
x^{2}+a x+2 b=0 \text{ and } x^{2}+2 b x+a=0
$$

both have real roots. Find the minimum value of $a^{2}+b^{2}$.
Explanation: Since the given equations all have real roots, we obtain the constraint conditions
$$
\left\{\begin{array}{l}
a^{2}-8 b \geqslant 0 \\
b^{2}-a \geqslant 0 \\
a>0, b>0
\end{array}\right.
$$

To find the minimum value of $a^{2}+b^{2}$, using algebraic methods is clearly challenging, so we consider a numerical and geometric transformation.

As shown in Figure 1, we draw the right half of the parabola $a^{2}=8 b(a>0)$, and then draw the upper half of the parabola $b^{2}=a(b>0)$. The intersection point of the two curves is $M(4,2)$. Thus, the point $P(a, b)$ that satisfies equations (1), (2), and (3) lies in the shaded region $G$ (including the boundary) shown in Figure 1.

Since $a^{2}+b^{2}$ represents the square of the distance from point $P(a, b)$ to the origin $O$, it is easy to see that when $a=4, b=2$,
$$
\left(a^{2}+b^{2}\right)_{\min }=|O M|^{2}=20 .
$$