Three, (25 points) Given that $a, b, c$ are positive integers, and the parabola $y=$ $a x^{2}+b x+c$ intersects the $x$-axis at two distinct points $A, B$. If the distances from $A, B$ to the origin are both less than 1, find the minimum value of $a+b+c$.

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Three, let the coordinates of $A, B$ be $\left(x_{1}, 0\right),\left(x_{2}, 0\right)$ and $x_{1}<0, x_{2}>0$,
\[
\begin{array}{l}
\therefore x_{1} x_{2}=\frac{c}{a}<0, \\
\therefore a c<0, \text { i.e., } a>0, c<0 .
\end{array}
\]
\[
\begin{array}{l}
\because x_{1}+x_{2}=-\frac{b}{a}>0, \\
\therefore b<0 .
\end{array}
\]
\[
\begin{array}{l}
\because \text { the parabola opens upwards, and } x_{1} x_{2}<0, \\
\therefore \text { the parabola intersects the } x \text{-axis at two points, } \\
\therefore \Delta=b^{2}-4 a c>0, \\
\because x_{1} x_{2}<0, \text { get } b>2 \sqrt{a c} .
\end{array}
\]
\[
\begin{array}{l}
\because|O A|=|x_{1}|,|O B|=|x_{2}|, \\
\therefore|O A| \cdot|O B|=|x_{1} x_{2}|=\left|\frac{c}{a}\right| \geqslant 1, \\
\therefore|c| \geqslant|a|, \text { and } a>0, c<0, \\
\therefore c \leqslant -a .
\end{array}
\]
\[
\begin{array}{l}
\because \text { the parabola passes through the point } (-1,1), \\
\therefore a(-1)^{2}+b(-1)+c>0 . \\
\text { get } b<2 \sqrt{a c}+1 \Rightarrow(\sqrt{a}-\sqrt{c})^{2}>1 .
\end{array}
\]
\[
\begin{array}{l}
\text { From (2) we get } \sqrt{a}-\sqrt{c}>1 . \\
\therefore \sqrt{a}>\sqrt{c}+1 .
\end{array}
\]
\[
\begin{array}{l}
\text { i.e., } a>(\sqrt{c}+1)^{2} \geqslant(\sqrt{1}+1)^{2}=4, \\
\therefore a \geqslant 5 .
\end{array}
\]
\[
\begin{array}{l}
\text { Also, } b>2 \sqrt{a c} \geqslant 2 \sqrt{5 \times 1}>4, \\
\therefore b \geqslant 5 .
\end{array}
\]
\[
\begin{array}{l}
\text { Taking } a=5, b=5, c=1, \text { the parabola } y=5 x^{2}+5 x+1 \text { satisfies the given conditions. } \\
\text { Hence, the minimum value of } a+b+c \text { is } 5+5+1=11 .
\end{array}
\]
(Provided by Liu Yuxuan)