Problem: If for every real x, $ax^2 + bx + c \geq 0 $ where a, b, c are reals such that $a < b$, what is the smallest value of $\frac{a+b+c}{b-a}$.
Let $$ L = \frac{a+b+c}{b-a} $$
I have a solution by two different other methods
1) By first using $b^2 - 4ac \leq 0$ and then substituting $b = a + k$, and do some algebraic manipulations, and use AM-GM in between to obtain $L \geq 3$
2) By substituting $\frac{b}{a} = t$, do some algebraic manipulations again, then use the fact that $\frac{b^2}{4a^2} = \frac{t^2}{4} \leq \frac{c}{a}$ to obtain a equation solely in $t$ and then just minimise it using calculus to again obtain $L \geq 3$
But I was trying to explore other methods to solve this problem and the following was my attempt at doing so,
Let us say the given is bigger than some constant $k$,
$$ L = \frac{a+b+c}{b-a} \geq k $$
But $f(1) = a + b + c$ and $f(0) = c$, thus $f(0) - f(-1) = c - (a - b + c) = b - a$ further,
Thus the given is equivalent to,
$$ \frac{f(1)}{c - f(-1)} \geq k$$
$$ \frac{f(0) - f(-1)}{f(1)} \leq \frac{1}{k} $$
I couldn't proceed any further using this method. Any suggestions on how to proceed using this, or any similar methods would be appreciated.
I figured another method, which is as follows:
$$ \frac{a + b + c}{b - a} = \frac{3f(1)}{f(1) - f(-2)} \geq 3 $$
Which follows since $f(1) \geq 0$ and the minima will be when $f(-2) = 0$