Given: Roots lie in $(0,1).$
Let $f(x)=ax^2-bx+c$ and it's roots be $\alpha$ and $\beta$
$\implies f(0) \times f(1) > 0$ (Can be easily verified from the parabolic graph of $f(x)$)
or $c(a-b+c)>0$
$\implies \frac{c}{a}(1-\frac{b}{a}+\frac{c}{a})>0$
$\implies \alpha \beta (1-(\alpha + \beta) + \alpha\beta) > 0$ (Using Vieta's Formula)
$\implies \alpha\beta(1-\alpha)(1-\beta) > 0$
Now,
Consider $\alpha(1-\alpha)$,
By AM-GM inequality,
$\alpha(1-\alpha)<\frac{1}{4}$
Similarly,
$\beta(1-\beta)< \frac{1}{4}$
By multiplying the above two inequalities, we get,
$\alpha\beta(1-\alpha)(1-\beta)<\frac{1}{16}$
$\implies \frac{c}{a}(1-\frac{b}{a}+\frac{c}{a})<\frac{1}{16}$
$\implies c(a-b+c)<\frac{a^2}{16}$
If we let $a$=$b$ and $c=1$, clearly we are getting the minimum value of $a$, i.e.
$\frac{a^2}{16}>1$
$a>4$ or minimum $a =5$
Since $D > 0$, we have $b^2-4ac > 0$ (where $D$ is the discriminant of $f(x)=0$)
this inequality is satisfied for $a=b=5$ which we calculated above
thus at $a=b=5$ and $c=1$ the minimum value of $abc=25$ is achieved.