Example 4 Given that $a, b, c$ are positive integers, and the quadratic equation $a x^{2}+b x+c=0$ has two real roots whose absolute values are both less than $\frac{1}{3}$. Find the minimum value of $a+b+c$.
(2005, National High School Mathematics League, Fujian Province Preliminary Contest)
Let $x_{1}$ and $x_{2}$ be the roots of the equation $a x^{2} + b x + c = 0$. By Vieta's formulas, we have
$$
x_{1} + x_{2} = -\frac{b}{a}, \quad x_{1} x_{2} = \frac{c}{a}.
$$
Thus, $x_{1}9$.
Therefore, the roots $x_{1}$ and $x_{2}$ of $a x^{2} + b x + c = 0$ are in the interval $\left(-\frac{1}{3}, 0\right)$. Hence, we can use the knowledge of the distribution of real roots of a quadratic equation to solve the problem.
Let $f(x) = a x^{2} + b x + c$. Then
$$
\left\{\begin{array}{l}
f(0) = c > 0, \\
f\left(-\frac{1}{3}\right) = \frac{1}{9} a - \frac{1}{3} b + c > 0, \\
-\frac{1}{3} < -\frac{b}{2a} < 0, \\
\Delta = b^{2} - 4ac > 0.
\end{array}\right.
$$
From the above, we have $-\frac{1}{3} < -\frac{b}{2a} < 0$, which implies $0 < \frac{b}{a} < \frac{2}{3}$, or $0 < b < \frac{2}{3} a$. Also, from $f\left(-\frac{1}{3}\right) > 0$, we get $\frac{1}{9} a - \frac{1}{3} b + c > 0$, which simplifies to $a - 3b + 9c > 0$, or $a + 9c > 3b$.
Since $a$, $b$, and $c$ are positive integers, $a + 9c$ and $3b$ are positive integers, so $a + 9c \geq 3b + 1$.
From the above, we have $a + 9c \geq 3b + 1 \geq 6 \sqrt{ac} + 1$.
Rearranging, we get
$$
(\sqrt{a} - 3 \sqrt{c})^{2} \geq 1.
$$

Combining this with $a > 9c$, we get $\sqrt{a} - 3 \sqrt{c} \geq 1$.
Thus, $\sqrt{a} \geq 1 + 3 \sqrt{c} \geq 1 + 3 \times 1 = 4$, which implies $a \geq 16$.
When $a = 16$, we have $c = 1$. In this case, $b \geq 2 \sqrt{ac} = 8$.
Upon verification, $16 x^{2} + 8 x + 1 = 0$ satisfies the given conditions. Therefore, the minimum value of $a + b + c$ is 25.
Note 1: The most common method for solving discrete extremum problems is to first estimate the range of the object of study, determine the possible extremum, and then construct to show that the extremum can be achieved.
Note 2: In [1], from $\Delta = b^{2} - 4ac \geq 0$, we get
$$
b^{2} \geq 4ac = 4 \times \frac{a}{c} \times c^{2} > 36.
$$