<s>[INST] Minimum value of $\dfrac{a+b+c}{b-a}$ 
$f(x)= ax^2 +bx +c ~ ~~(a<b)$ and $f(x)\ge 0~ \forall x \in \mathbb R$ .
  Find the minimum value of $\dfrac{a+b+c}{b-a}$

Attempt: 
$b^2 \le 4ac$
$f(1) = a+b+c$
$f(0) = c$
$f(-1) = a-b+c$
$a>0$
and $c>0$
I am unable to utilize these things to find the minimum value of the expression $\equiv \dfrac{a+b+\frac {b^2}{4a}}{b-a}$
The answer given is $3$. 
 [/INST] Rewrite the equation as $x^2+Bx+C$. Now, $B^2\le 4C$ So, the minimum value of $C=\dfrac{B^2}{4}$. Now, $\dfrac {a+b+c}{b−a} =\dfrac {1+B+(B^2/4)}{B−1}$. Find the minimum value of function. 
</s>