Find the minimum value of $\frac{a+b+c}{b-a}$

As you noted, we have $b > a > 0$, and $b^2-4ac\le 0$, hence $c\ge {\large{\frac{b^2}{4a}}}$.

Letting $t=b-a$, we have $t > 0$, and $b=a+t$. \begin{align*} \text{Then}\;\;&\frac{a+b+c}{b-a}\\[4pt] &\ge \frac{a+b+\frac{b^2}{4a}}{b-a}\\[4pt] &=\frac{(2a+b)^2}{4a(b-a)}\\[4pt] &=\frac{(3a+t)^2}{4at}\\[4pt] &=\frac{9a^2+6at+t^2}{4at}\\[4pt] &=\frac{9a}{4t}+\frac{3}{2}+\frac{t}{4a}\\[4pt] &=\frac{3}{2}+\left(\frac{9a}{4t}+\frac{t}{4a}\right)\\[4pt] &\ge \frac{3}{2}+2\sqrt{\frac{9}{16}}\qquad\text{[by $\text{AM-GM}$]}\\[4pt] &=3\\[4pt] \end{align*} so $3$ is a lower bound.

To show that $3$ is realizable, we can use $a=1$, and $t=3$ (which makes the $\text{AM-GM}$ inequality an equality), so $b=a+t=4$, and finally, letting $c={\large{\frac{b^2}{4a}}}=4$, we get $$\frac{a+b+c}{b-a}=3$$ which gives the minimum possible value.

Hint: complete the square. Write $ax^2+bx+c=a(x-d)^2+e$ where you can express $d,e$ in terms of $a,b,c$. You need $a,e \gt 0$ but $d$ can be anything.

If $f(x) \ge 0\to b^2-4ac \le 0$ then forming the lagrangian

$$ L(a,b,c,\lambda,\epsilon) =\phi(a,b,c) +\lambda(b^2-4ac+\epsilon^2) $$

with

$$ \phi(a,b,c) = \frac{a+b+c}{b-a} $$

The stationary points are computed by solving

$$ \nabla L = \left\{ \begin{array}{rcl} \frac{a+b+c}{(b-a)^2}-4 c \lambda +\frac{1}{b-a}=0 \\ -\frac{a+b+c}{(b-a)^2}+2 b \lambda +\frac{1}{b-a}=0 \\ \frac{1}{b-a}-4 a \lambda =0 \\ b^2+\epsilon ^2-4 a c=0 \\ 2 \epsilon \lambda =0 \\ \end{array} \right. $$

we obtain

$$ \left[ \begin{array}{cccccc} a & b & c & \lambda & \epsilon & \phi\\ -\frac{b}{2} & b & -\frac{b}{2} & -\frac{1}{3 b^2} & 0 & 0 \\ \frac{b}{4} & b & b & \frac{4}{3 b^2} & 0 & 3 \\ \end{array} \right] $$

hence the feasible solution is

$$ a = \frac b4, c = b $$

giving a minimum value of $3$