We will work with $f(x)=ax^2-bx+c$ with $a,b,c>0$ We have $0<\frac{c}{a}<\frac{1}{1000000}$ and $0<\frac{b}{a}<\frac{1}{500} \to a>500b$ So $a>1000000c>0$ If $c=1$ then we have $a>1000000$ and $b^2-4a>0 \to 2000b<4a<b^2 \to b>2000 \to b \geq 2001$ we have $\frac{b+\sqrt{b^2-4a}}{2a} \leq \frac{1}{1000}$ $b^2-4a \leq (\frac{a}{500}-b)^2$ $\frac{a^2}{250000}-\frac{ab}{250}+4a \geq 0$ $a \geq 1000b-1000000 \geq 1001000$ And for $a=1001000$ we can take $f(x)=1001000x^2-2001x+1=(1001x-1)(1000x-1)$ If $c \geq 2$ then $a>2000000$ and can`t be minimal