Example 7 Given $a, b, c \in \mathbf{N}_{+}$, and the parabola $f(x) = ax^{2} + bx + c$ intersects the $x$-axis at two different points $A$ and $B$. If the distances from $A$ and $B$ to the origin are both less than 1, find the minimum value of $a + b + c$.
(1996, National Junior High School Mathematics Competition)
Let $f(x)=a x^{2}+b x+c, A\left(x_{1}, 0\right)$, and $B\left(x_{2}, 0\right)$. According to the problem, we have
$$
\left\{\begin{array}{l}
\Delta=b^{2}-4 a c>0, \\
f(1)=a+b+c>0, \\
f(-1)=a-b+c>0 .
\end{array}\right.
$$

From (1), we get $b>2 \sqrt{a c}$; from (3), we get $a+c>b$.
$$
\begin{array}{l}
a+c>b \Rightarrow a+c \geqslant b+1>2 \sqrt{a c}+1 \\
\Rightarrow(\sqrt{a}-\sqrt{c})^{2}>1 .
\end{array}
$$

Since $\frac{c}{a}=x_{1} x_{2}c$, we have
$$
\begin{array}{l}
\sqrt{a}>\sqrt{c}+1 \\
\Rightarrow a>(\sqrt{c}+1)^{2} \geqslant(1+1)^{2}=4 .
\end{array}
$$

Thus, $a \geqslant 5, b>2 \sqrt{5}>4$.
So, $b \geqslant 5$.
Therefore, $a+b+c \geqslant 11$.
Taking $a=5, b=5, c=1$, we get
$$
\begin{array}{l}
f(x)=5 x^{2}+5 x+1=0, \\
-1<x_{1,2}=\frac{-5 \pm \sqrt{5}}{10}<1
\end{array}
$$

which satisfies the conditions. Therefore, the minimum value of $a+b+c$ is 11.
The above text illustrates the method for solving the distribution of roots of a quadratic equation. Of course, for a specific problem, there can be multiple methods to solve it.