<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] Alt. hint: $\;f(-2) = 4a - 2 b + c \ge 0 \iff 2a + c \ge 2(b-a)\,$, then:
$$
\dfrac{a+b+c}{b-a} = 1 + \frac{2a+c}{b-a} \;\ge\; 1 + 2 = 3
$$
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