Quadratic equation, math olympiad question So this is a 9-10th grade, math olympiad problem I found. Define the parabola $y=ax^2+bx+c$ such that $a,b,c$ are positive integers. Suppose that the roots of the quadratic equation $ax^2+bx+c=0$ are $x_1,x_2$ such that $|x_1|,|x_2|>1$, then compute the minimum value of $abc$ and when the minimum occur what is the value of $a+b+c$. Maybe there is a way to solve this using advanced math, calculus, but I am sure there is an elementary way of solving this without using any calculus. 
 A: $f(x) = ax^2+bx+c$ cannot have any positive root when $a$, $b$, $c$ are all positive. So we have $x_1, x_2<-1$.
The midpoint between the roots is $-b/2a$. That must be less than $-1$, so we have $$ \tag{1} \frac{-b}{2a}<-1 \implies \frac{b}{2a}>1 \implies b>2a $$
Furthermore the larger root is less than $-1$ too, so
$$ \tag{2} f(-1) >0 \implies a-b+c>0 \implies c > b-a $$
Given these conditions, the smallest values $a$, $b$ and $c$ can have are $a=1$, $b=c=3$. Unfortunately $x^2+3x+3$ doesn't have roots at all. Increasing $c$ can't help, so $b$ must be at least $4$, and so $c=4$ too.
$x^2+4x+4$ has a double root at $x_1=x_2=-2$ and if that is allowed by the conditions we have $abc=16$ and $a+b+c=9$.
If $a$ was larger, say $a=2$ conditions $(1)$ and $(2)$ would give $b\ge 5$ and $c\ge 4$, which would give $abc\ge40$, which is much larger so we can discount that.

If the double root is not allowed, we would need to go to higher $b$ still, so consider $b=5$ -- then we get $c=5$ too. Now at last there are two different roots (we don't need to compute them, only observe that the discriminant is positive). The product $abc$ is now $25$ which is still less than $40$, and the answer for $a+b+c$ is then $11$.
