Let $f(x)= ax^2 +bx+c$, such that $f(x)\geq0$. Find minimum value of $\frac{f(1)}{f(0)-f(-1)}.$
My attempt: denominator should be maximum so $f(-1)=0$ as $f(x)$ is always greater than or equal to zero. So I got $b=2a$, $c=a$. But then I am getting the answer as $4$ whereas the answer is $3$.
Wlog we can assume $a=1$. So the task becomes to minimize $F(b,c)=\frac{1+b+c}{b-1 }$ with $b^2\leq4c$. Let us start by taking the derivative with respect to $c$, $\frac{\partial F}{\partial c}=\frac{1}{b-1}=0$. We see that we have no extremal points. So we turn to the boundary $F(b,\tfrac{b^2}{4})$. Again looking for extremal points we differentiate this time with respect to $b$ and end up with $\frac{b^2-2b-8}{4(b-1)^2}=0$ which implies $b\in\{-2,4\}$. Inserting yields $F(-2,1)=0$ and $F(4,4)=3$.