Distinct roots of $ax^2-bx+c=0$ in $(0,1)$, where $a,b,c\in \mathbb{Z}^+$ 
Question Statement:-
Let $a,b,c$ be positive integers and consider all the quadratic equations of the form $ax^2-bx+c=0$ which have two distinct real roots in $(0,1)$. Find the least positive integers $a$ and $b$ for which such a quadratic equation exist.


Attempt at a solution:-
Let $f(x)=ax^2-bx+c$. Now as the roots of $f(x)=0$ lie in $(0,1)$ hence $f(0)\cdot f(1)\gt0$.
$$D\gt0\implies b^2-4ac\gt0$$
$$f(0)=c \\ f(1)=a-b+c$$
As, $a,b,c\in\mathbb{Z}^+$, so
$$f(0)\ge1 \implies c\ge1\tag{1}$$
$$f(1)\ge1\implies a-b+c\ge1\tag{2}$$
$$f(0)\cdot f(1)\ge1\implies c(a-b+c)\ge1\implies c^2+(a-b)c-1\ge0\tag{3}$$
Now, for the quadratic inequality in $c$ that we obtained in $(3)$, for it to hold we get
$$D\le0\implies (a-b)^2+4\le0$$
The above inequality doesn't hold $\forall a,b \in \mathbb{Z}^+$.
After arriving at this conclusion I am kind of stuck on what to conclude from the above result or rather am I going in the right direction.
A little push in the right direction in the form of a hint or rather what should be the line of thought for attempting the question. Please don't write the solution, just the line of thought to arrive at the solution would be enough.
 A: The brute force approach works out fairly quickly in this case.


*

*$f(0) \gt 0$ $\implies$ $c \gt 0$ and $f(1) \gt 0$ $\implies$ $b \lt a + c$

*the product of the roots must be in $(0,1)$ $\implies$ $c < a$

*the discriminant $\Delta \gt 0$ $\implies$ $b^2 \gt 4 ac$
It follows that $c \ge 1$, $a \ge c+1$ and $b \in (\sqrt{4ac}, a+c)$. Trying the first few values starting at the lowest possible $c=1$ and $a=2$ gives:
$$
\begin{align}
& c = 1 & a = 2 \quad \implies \quad & b \in (\sqrt{8},3) & = & \;\emptyset \\
& & a = 3 \quad \implies \quad & b \in (\sqrt{12},4) & = & \;\emptyset \\
& & a = 4 \quad \implies \quad & b \in (\sqrt{16},5) & = & \;\emptyset \\
& & a = 5 \quad \implies \quad & b \in (\sqrt{20},6) & = & \;\{\;5\;\}
\end {align}
$$
The latter gives the solution $5 x^2 - 5x +1$ which can be easily verified to have both roots in $(0,1)$.

[ EDIT ]   To complete the proof that $a=5$ is indeed the smallest possible value, the cases $c=2,3$ must also be checked (we can stop at $3$ since $c \ge 4$ implies $a \ge 5$).

$$
\begin{align}
& c = 2 & a = 3 \quad \implies \quad & b \in (\sqrt{24},5) & = & \;\emptyset \\
& & a = 4 \quad \implies \quad & b \in (\sqrt{32},6) & = & \;\emptyset
\end {align}
$$
$$
\begin{align}
& c = 3 & a = 4 \quad \implies \quad & b \in (\sqrt{48},7) & = & \;\emptyset
\end {align}
$$
A: After a thinking a lot on the question I was able to come up with a good enough solution.
Let $f(x)=ax^2-bx+c$. Let the roots of the equation $f(x)=0$ be $\alpha$ and $\beta$. Then, $\alpha,\beta\in(0,1)$.
As the roots of the equation $f(x)=0$ are real and distinct, then $b^2\gt4ac$.Also, due to the roots lying in $(0,1)$, hence $f(0)\gt0$, $f(1)\gt0$ and $f(0)f(1)\gt0$
Now, $$f(0)=c\ge1\tag{1}$$ 
$$f(1)=a-b+c\ge1\tag{2}$$
$$\therefore \qquad f(0)\cdot f(1)\ge1\qquad\qquad\tag{3}$$
If $\alpha$ and $\beta$ are the zeroes of $f(x)$, then $f(x)=a(x-\alpha)(x-\beta)$
$$\therefore f(0)\cdot f(1)=a\alpha(1-\alpha)a\beta(1-\beta)\ge 1$$
Using A.M-G.M inequality we get,
$$\dfrac{b+(1-b)}{2}\ge\sqrt{b(1-b)}\implies b(1-b)\le\dfrac{1}{4}$$
equality occurs when $b=\dfrac{1}{2}$
$$\therefore \qquad\qquad\alpha(1-\alpha)\le\dfrac{1}{4}\qquad\qquad \beta(1-\beta)\le\dfrac{1}{4}$$
In both the cases equality occurs when $\alpha=\beta=\dfrac{1}{2}$, but as $\alpha\neq\beta$
So, $$\alpha(1-\alpha)\beta(1-\beta)\lt\dfrac{1}{16}\implies a\alpha(1-\alpha)a\beta(1-\beta)\lt\dfrac{a^2}{16}\\
\implies \dfrac{a^2}{16}\gt f(0)\cdot f(1)\implies |a|\gt4$$
$$\therefore a_{min}=5$$
Also, $$\because \qquad\qquad\alpha\cdot\beta\lt1\implies c\lt a_{min}\qquad\qquad\tag{4}$$
$$(2)\implies a+c\ge1+b\implies a_{min}+c-1\ge b\implies 4+c\ge b \\
\implies c^2+8c+16\ge b^2\implies c^2+8c+16 \gt 4a_{min}c\\
\implies c^2-12c+16\gt0\implies c\in(0,6-2\sqrt5)\cup(6+2\sqrt5,\infty)$$
But, as $c\lt a_{min}$, hence $c=1$
Now, from $$b^2\gt 4a_{min}c\implies b^2\gt20$$ 
Hence, $b_{min}=5$
So, the value of least positive integers $a$,$b$ and $c$ for which $f(x)$ has roots b/w $0$ and $1$ is $(a,b,c)=(5,5,1)$
Other answers are still welcome.
A: Let $x_1, x_2$ be the two roots with $0\lt x_1\lt x_2\lt 1$. The graph of $f(x)=ax^2-bx+c$ is a parabola (with $a\gt 0$) whose vertex $V$ is given by $f\space'(x)=2ax-b=0$, i.e. $$V=\left(\dfrac{b}{2a},f\left(\dfrac{b}{2a}\right)\right)=\left(\dfrac{b}{2a},\dfrac{4ac-b^2}{4a}\right)$$ Because the roots are real we have $b^2-4ac\gt 0$, furthermore  $f(x)$ is decreasing for $x\lt\dfrac{b}{2a}$ and increasing for $x\gt\dfrac{b}{2a}$ so one has that both $f(0)$ and $f(1)$ are positive (since $f(0)\le0\Rightarrow x_1\lt0$ and $f(1)\le 0\rightarrow x_2\gt1$) It follows we have the conditions
$$\begin{cases}f(1)=0\iff a+c\gt b\\4ac\lt b^2\\f(0)=c\gt 0\end{cases}$$ we can use just $$\begin{cases}a+c\gt b\\4ac\lt b^2\end{cases}$$ from which we get directly $b\ge 3$. Taking the minimun value $c=1$ we find incompatibility for $b=3$ and $b=4$ but for $b=5$ we have
$$\begin{cases}a+1\gt b\\4a\lt b^2\end{cases}\Rightarrow 4\lt a\lt\frac{25}{4}\Rightarrow 4\lt a\le 6$$ so have solutions $$(a,b,c)=(5,5,1),(6,5,1)$$  and it is easily verified that the equations $5x^2-5x+1=0$ and $6x^2-5x+1=0$ have roots according to the proposed problem.
Thus the required minimun values of $(a,b)$ is $\color{red}{(5,5)}$.
It is a simple question to verify that for values of $c\gt1$ there are not solution for $b\lt5$, say $b=4$, because we would deduce $a\gt 3-h$ and $a\lt \dfrac{4}{1+h}\le 2$
