If a, b, c are positive integers and the roots of the equation $a{x}^2-bx+c=0$lie in $(0,1)$ and distinct, then

$\alpha ,\beta \in (0,1)$ on applying A.M –G.M . inequality to $\alpha ,1-\alpha$ and also to $\beta$

$1-\beta \frac{\alpha +1-\alpha }{2}\ge \sqrt{\alpha (1-\alpha )}$

$\frac{\beta +1-\beta }{2}\ge \sqrt{\beta (1-\beta }\Rightarrow 0(\alpha 0-\alpha )<\frac{2}{4}$and 

$0<\beta (1-\beta )<\frac{1}{4}\Rightarrow \alpha \beta (1-\alpha )(1-\beta )<\frac{1}{16}$

Since a, b, c are positive intergers $c(a-b+c)$ is an integer 

Let $f\left(x\right),f\left(0\right)$ have same sign and $f\left(0\right)f\left(1\right)$

$\ge 1\Rightarrow 1\le f\left(0\right)f\left(1\right)$

$\Rightarrow 1\le a^2\alpha \beta (1-\beta )(1-\beta )<a^2\left(\frac{1}{16}\right)\Rightarrow a^2>16$

$\Rightarrow a\ge 5$since $a{x}^2-bx+c=0$ are real and distinct there fore 

$b^2>4ab\Rightarrow b^{2}20c\Rightarrow b^2>20\Rightarrow b\ge 5$

$\therefore$Minimum value of b c & 5

( since $c\in N$ be have $c\ge 1$

$\therefore$ Minimum value of c is 1;b is 5; a is 5