Example 13 Given that $a$, $b$, and $c$ are all positive integers, and the parabola $y=a x^{2}+b x+c$ intersects the $x$-axis at two distinct 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 League)
Solution: Let $A\left(x_{1}, 0\right)$ and $B\left(x_{2}, 0\right)$, and $x_{1} < 0 < x_{2}$,  
then $x_{1} < 0, \\ 
\therefore b > 2 \sqrt{a c} . \\ 
\text{Also, } \because |O A| = |x_{1}| > 1$. Therefore, the parabola opens upwards, and when $x = -1$, $y > 0$, so $a(-1)^{2} + b(-1) + c > 0$, which means $b < a + c + 1$. Since $b > 2 \sqrt{a c}$, we have $2 \sqrt{a c} + 1 < a + c$,  
i.e., $(\sqrt{a} - \sqrt{c})^{2} > 1$.  
From equation (2), we get $\sqrt{a} - \sqrt{c} > 1$,  

i.e., $\sqrt{a} > \sqrt{c} + 1$.  
Thus, $a > (\sqrt{c} + 1)^{2} \geq (\sqrt{1} + 1)^{2} = 4$.  
Solving, we get $a \geq 5$.  
$$
\begin{array}{l}
\text{Also, } \because b > 2 \sqrt{a c} \geq 2 \sqrt{5 \times 1} > 4, \\
\therefore b \geq 5 .
\end{array}
$$

When $a = 5, b = 5, c = 1$, the parabola $y = 5 x^{2} + 5 x + 1$ satisfies the given conditions.  
Therefore, the minimum value of $a + b + c$ is 11.