Minimum values of coefficients of a quadratic Given a function $f(x)= ax^{2} + bx + c$ where $a<b$ and $f(x)\geq{0}$ for all real values of x. Then how would one find the minimum value of the relation between coefficients of the give quadratic. For ex, 
How would one find the min value of $\frac{a+b+c}{b-a}$.
my work so far 
I concluded that $\frac{a+b+c}{b-a}$ is the same as $\frac{f(1)}{b-a}$ and as per the given conditions $b^2 -4ac\leq{0}$ and I tried finding some triplets of $a,b, c$ and find the minimum value by observation but had no luck. 
All help is greatly appreciated

Let $f(x)=ax^2+bx+c$ where
$$\forall x \in \mathbb R, f(x) \ge 0 \qquad
\text{and} \qquad a < b \tag A$$
$$\text{Find $\min \dfrac{a+b+c}{b-a}$} \tag{B}$$
If $a=0$, then the problem is ill-defined. When $a < 0$, then $\displaystyle \lim_{x \to \infty}f(x) \to -\infty$. So $a > 0$.
Since $f(0) = c$, then we must have $ c \ge 0$, but we can do better than that. Clearly $f(x)$ is a parabola above or on the $x$-axis. So, holding $a$ and $b$ fixed, the value of $c$ that will minimize $\dfrac{a+b+c}{b-a}$ is the value, $c = -\frac 14b^2$, that puts the vertex of $f(x)$ on the $x$-axis. Combining that with $c \ge 0$, we get $c=0$. Since we require $a < b$, we let $b = aN$ where 
$N > 0$.
We get $f(x) = a(x^2 + Nx + 0)$ and we want to minimize 
\begin{align}
   \dfrac{a+b+c}{b-a} 
   &= \dfrac{a+aN+0}{aN-a} \\
   &= \dfrac{1+N}{N-1} \\
   &= 1 + \dfrac{2}{N-1}
\end{align}
which approaches $1$ as $N$ approaches infinity.
For example, $f(x) = x^2 + Nx + 0$, where $N > 1$ satisfies condition (A) and
$\dfrac{a+b+c}{b-a} =  \dfrac{N+1}{N-1} \to 1$ as $N \to \infty$.
