If the vertex of the parabola $$y=ax^2+bx+c \qquad (0<2a<-b)$$is not below the x-axis. Let there be three points on the parabola: $A(-1,p), B(0,q)$ and $C(1,r)$. Then what is the minimum value of $\dfrac{p}{q-r}$ ?
Hint: the answer is $3$.
My thoughts:
Since the axis of symmetry of the function is $x=-\dfrac{b}{2a}>1$ and $\Delta=b^2-4ac<0$, $\dfrac{p}{q-r}=-\dfrac{a-b+c}{a+b}$, I thought of solving it by the image of the function. But it seems difficult to find the answer only through the limited conditions.
Let $f(x)=a(x-x_s)^2+y_s$; we know that $a>0$, $x_s\ge1$ and $y_s\ge0$. We search for a minimum of $$\begin{align} d(a,x_s,y_s)&:=\frac{f(-1)}{f(0)-f(1)}\\ &=\frac{(1+x_s)^2}{2x_s-1}+\frac{y_s}{a(2x_s-1)}\\ &=d(a,x_s,0)+\frac{y_s}{a(2x_s-1)}. \end{align}$$
First verify that for $y_s=0$ the minimum of $$d(a,x_s,0)=\frac{(1+x_s)^2}{2x_s-1}$$ is achieved iff $x_s=2$; then $d=3$, it doesn't depend on $a$.
Finally $$d(a,x_s,y_s)-d(a,x_s,0)=\frac{y_s}{a(2x_s-1)}>0.$$