<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] Actually, you are on the right track! Now let's complete it with only elementary calculus.
$$\frac{a+b+\frac{b^2}{4a}}{b-a}=\frac{4a^2+4ab+b^2}{4a(b-a)}=\frac{(2a+b)^2}{4a(b-a)}$$
Knowing that $b>a$, let's set $b=ka, k>1$.
$$\frac{(2a+b)^2}{4a(b-a)}=\frac{(2+k)^2a^2}{4(k-1)a^2}=\frac{(2+k)^2}{4(k-1)}$$
Now you just need to find the minimum of this function for $k>1$. You can differentiate it and check that $k=4$ is a critical point, and you get $\frac{a+b+c}{b-a}=3$ there as a minimum.
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