Find the minimum value of $\frac{a+b+c}{b-a}$ Let $f(x)=ax^2+bx+c$ where $(a<b)$ and $f(x)\geq 0$  $\forall x\in R$. 
Find the minimum value of $$\frac{a+b+c}{b-a}$$
If $f(x)\geq 0$  $\forall x\in R$ then $b>a>0$ and  $b^2-4ac\leq 0$ implying that $c>0$. After this not able to find way out.
 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.
A: 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.
A: 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 $3$
A: Given $f(x) = ax^2+bx+c\geq 0\forall x \in \mathbb{R}$
Now put $x=-2,$ We get $f(-2)\geq 4a-2b+c\geq 0\Rightarrow 2a+c\geq 2(b-a)$
So $\displaystyle \frac{a+b+c}{b-a}=1+\frac{2a+c}{b-a}\geq 1+2=3.$
