<s>[INST] Prove that $\frac{a + b + c}{b - a} > 3$ with $ab < b^2 < 4ca$. 
$a$, $b$, $c$ are three positives such that $ab < b^2 < 4ca$. Prove that $$\large \dfrac{a + b + c}{b - a} > 3.$$

I can't think of a way to get around this problem. Although I can see that based on the given condition, $ax^2 + bx + c = 0$ has no roots, which adds almost no information whatsoever.
If you have written an answer below, thanks for that!
 [/INST] Assuming $f(x) = ax^2+bx+c>0$ for all real $x$ and
$a>0$ and $(b-a)>0$
put $x=-2,$ We get 
$f(-2)
=4a-2b+c> 0\Rightarrow 2a+c>2(b-a)$
$\Rightarrow \displaystyle \frac{a+b+c}{b-a}=1+\frac{2a+c}{b-a}>1+2=3.$
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