Find minimum possible values of the leading coefficient of quadratic equation let the polynomial $$f(x) = ax^2- bx+ c$$ (where a, b, are positive integers). If $f(p)=f(q)=0$ where $0<p<q<1$ then what is the minimum possible value of a.
I used some basic inequalities to get the following - 
$$1) b<2a$$
$$2) a>c$$
$$3) a+c>b$$
But couldn't reach for the inequality for a.  
Edit1- The answer is 5

You are on the right track. First of all, you don't really need condition $1)$ since it is implied by $2)$ and $3)$. On the other hand, you need to add condition $b^2>4ac$, for example $2x^2-x+1$ satisfies your conditions but has no real roots.
Now, you can exploit the fact that $a,b,c$ are positive integers to get better bounds:
\begin{align}
a&\geq c + 1\\
a+c&\geq b + 1\\
b^2&\geq 4ac + 1
\end{align}
which can be rewritten as $$\max\{b-c+1,c+1\}\leq a\leq \frac{b^2-1}{4c}.$$
Since $\frac{b^2-1}{4c}\geq c + 1$, we get that $b\geq 2c + 1$ and from there that $a\geq b - c + 1\geq c + 2$. 
Iterate the procedure, i.e. $\frac{b^2-1}{4c}\geq c + 2$ implies $$b\geq \sqrt {4c^2+8c + 1} > 2c + 1$$ and again, since we are working with integers, $b\geq 2c + 2$, or $a\geq b - c + 1\geq c + 3$.
Rinse and repeat, $\frac{b^2-1}{4c}\geq c + 3$ implies $$b\geq \sqrt{4c^2+12c + 1}> 2c + 2,$$
i.e. $b\geq 2c + 3$ which implies $a\geq b - c + 1\geq c + 4$.
Notice that repeating the same thing will not give us $$b\geq \sqrt{4c^2+16c + 1} > 2c + 3$$ so we cannot improve the bound $a\geq c + 4$. But, this bound gives us plenty, since it implies that $a\geq 5$.
Finally, to see that $a = 5$ works, you just need to find an example, for instance, $5x^2 -5x + 1$ gives roots $x_{1,2} = \frac{1}{10}(5\pm\sqrt 5)$.
