TITLE: What is the minimum value of $abc$
QUESTION [6 upvotes]: If the roots of the equation $$ax^2-bx+c=0$$ lie in the interval $(0,1)$, find the minimum possible value of $abc$.   
Edit: I forgot to mention in the question that $a$, $b$, and $c$ are natural numbers. Sorry for the inconvenience.
Edit 2: As  Hagen von Eitzen said about the double roots not allowed, I forgot to mention that too. Extremely sorry :(

I tried to use $D > 0$, where $D$ is the discriminant but I can't further analyze in terms of the coefficients. Thanks in advance!

REPLY [8 votes]: The discrimimnat $D=b^2-4ac$ must be positive to ensure two distinct real roots.
(If double root is not forbidden, we have $4x^2-4x+1$ with double root at $\frac12$ and $abc=16$).
Next, we must have $f(1)>0$, i.e. $$a+c>b.$$
For naturals $a,c$ we also have $ a+c\le 1+ac$ and conclude $$\tag1b\le ac.$$
If $b\le 4$ we obtain $b\le ac<\frac14b^2\le b$, contradiction. (NB: If we relax the condition that the roots be distinct, the $<$ becomes a $\le$ and instead of a contradiction we find $b=ac=4$, hence $abc=16$).
Hence $b\ge 5$ and by $(1)$
$$abc\ge b^2\ge 25.$$
The minimum is indeed attained as can be seen by making all iniequalities sharp, which gives: Either $(a,b,c)=(5,5,1)$ or $(a,b,c)=(1,5,5)$. The first of these indeed gives two roots in $(0,1)$.