Example 9 Let $a, b, c$ be 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
Let the two real roots of the equation be $x_{1}$ and $x_{2}$. By Vieta's formulas, we know $x_{1}x_{2} = \frac{c}{a} = 9$. Therefore,
\[ b^{2} \geqslant 4ac = 4 \times \frac{a}{c} \times c^{2} > 4 \times 9 \times 1^{2} = 36. \]
Thus, $b \geqslant 7$.
Also, $\frac{b}{a} = (-x_{1}) + (-x_{2}) \Rightarrow \frac{3}{2} b \geqslant \frac{21}{2}$, which implies $a \geqslant 11$.
(1) When $b = 7$, from $4a \leqslant 4ac \leqslant b^{2}$ and $a \geqslant 11$, we know $a = 11$ or $12$, and $c = 1$. However, the equation $11x^{2} + 7x + 1 = 0$ has roots $-\frac{7 + \sqrt{5}}{22} - \frac{1}{3}$, which gives
\[ 4 + \sqrt{16 - a} > 0. \]
Since $f(15) = 0$, $a$ can only be 16.
At this point, $a + b + c = 25$, and the equation $16x^{2} + 8x + 1 = 0$ has roots $x_{1} = x_{2} = -\frac{1}{4}$, which satisfies the problem's conditions.
(3) When $b \geqslant 9$, $a > \frac{3}{2}b \geqslant \frac{27}{2}$, so $a \geqslant 14$. Therefore, $a + b + c \geqslant 14 + 9 + 1 = 24$.

If $a + b + c < 25$, then it can only be $a = 14$, $b = 9$, and $c = 1$. In this case, the equation $14x^{2} + 9x + 1 = 0$ has roots $-\frac{1}{2}$ and $-\frac{1}{7}$, which does not meet the problem's conditions.
Thus, $a + b + c \geqslant 25$.
In summary, the minimum value of $a + b + c$ is 25.