<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] If $f(x) \ge 0\to b^2-4ac \le 0$ then forming the lagrangian
$$
L(a,b,c,\lambda,\epsilon) =\phi(a,b,c) +\lambda(b^2-4ac+\epsilon^2)
$$
with 
$$
\phi(a,b,c) = \frac{a+b+c}{b-a}
$$
The stationary points are computed by solving
$$
\nabla L = \left\{
\begin{array}{rcl}
 \frac{a+b+c}{(b-a)^2}-4 c \lambda +\frac{1}{b-a}=0 \\
 -\frac{a+b+c}{(b-a)^2}+2 b \lambda +\frac{1}{b-a}=0 \\
 \frac{1}{b-a}-4 a \lambda =0 \\
 b^2+\epsilon ^2-4 a c=0 \\
 2 \epsilon  \lambda =0 \\
\end{array}
\right.
$$
we obtain
$$
\left[
\begin{array}{cccccc}
a & b & c & \lambda & \epsilon & \phi\\
 -\frac{b}{2} & b & -\frac{b}{2} & -\frac{1}{3 b^2} & 0 & 0 \\
 \frac{b}{4} & b & b & \frac{4}{3 b^2} & 0 & 3 \\
\end{array}
\right]
$$
hence the feasible solution is
$$
a = \frac b4, c = b
$$
giving a minimum value of $3$
</s>