<s>[INST] 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.
 [/INST] 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.
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