<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] 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 $\displaystyle \frac{a+b+c}{b-a}=1+\frac{2a+c}{b-a}\geq 1+2=3.$
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