One, (20 points) In the quadratic function $f(x)=a x^{2}+b x$ $+c$, $a$ is a positive integer, $a+b+c \geqslant 1, c \geqslant 1$, and the equation $a x^{2}+b x+c=0$ has two distinct positive roots less than 1. Find the minimum value of $a$.

保留源文本的换行和格式，直接输出翻译结果如下：

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One, (20 points) In the quadratic function $f(x)=a x^{2}+b x$ $+c$, $a$ is a positive integer, $a+b+c \geqslant 1, c \geqslant 1$, and the equation $a x^{2}+b x+c=0$ has two distinct positive roots less than 1. Find the minimum value of $a$.
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Given the equation $a x^{2}+b x+c=0$ has two roots $x_{1} 、 x_{2}(0<x_{1}<x_{2})$, and $x_{1} x_{2}=\frac{c}{a}=\frac{3}{5}$, we have $a x_{1} x_{2}=3$.

Since $x_{1}+x_{2}=\frac{11}{10}$, we have $a(x_{1}+x_{2})=\frac{11}{10}a>4.4$.

Given $a>4$, and $a$ is a positive integer, so, $a \geqslant 5$.

Taking $f(x)=5\left(x-\frac{1}{2}\right)\left(x-\frac{3}{5}\right)=5 x^{2}-\frac{11}{2} x+ \frac{3}{2}$, which satisfies the conditions, hence the minimum value of $a$ is 5.