No Fibbing 2

$\frac {G_{x+1}}{G_x} = \frac {G_x + (F_{x+1}F_{x}F_{x-1})}{G_x}$ = $1 + \frac {F_{x+1}F_{x}F_{x-1}}{G_x}$

Using the relationship $\phi = \lim_{n \to \infty}\ \frac {F_{n+1}}{F_n} = \frac {1 + \sqrt{5}}{2}$ we know that $\lim_{x \to \infty}$ $F_{x+1}F_{x}F_{x-1}$ = $\phi ^3 F_{x}F_{x-1}F_{x-2}$

So $\lim_{x \to \infty}$ $\frac {G_{x+1}}{G_x} = 1 + \frac {\phi^3 F_{x}F_{x-1}F_{x-2}}{G_x}$ $\Rightarrow Equ$ $1$

$\lim_{x \to \infty}$ $\frac {G_{x+1}}{G_x} \equiv \frac {G_{x}}{G_{x-1}}$

$\frac {G_{x}}{G_{x-1}}$ = $\frac {G_{x}}{G_{x} - (F_{x}F_{x-1}F_{x-2})}$

$\lim_{x \to \infty}$ $1 + \frac {\phi ^3 (F_{x}F_{x-1}F_{x-2})}{G_x}$ = $\frac {G_{x}}{G_{x} - (F_{x}F_{x-1}F_{x-2})}$ $\Rightarrow Equ$ $2$

Let $F_{x}F_{x-1}F_{x-2} = A$

Using the fact that $Equ$ $1$ $= Equ$ $2$ , $\lim_{x \to \infty}$ $1 + \frac {\phi^3 A}{G_x} = \frac {G_x}{G_{x} - A}$ and $\frac {G_x}{G_{x} - A} = 1 + \frac {A}{G_{x} - A}$

$\lim_{x \to \infty}$ $\frac {\phi^3 A}{G_x} = \frac {A}{G_{x} - A}$

$\lim_{x \to \infty}$ $G_x = - \frac {\phi ^3 A^2}{A- \phi ^3 A}$

Substituting in to $1 + \frac {\phi^3 A}{G_x}$ : $1 + \frac {\phi^3 A}{- \frac {\phi ^3 A^2}{A- \phi ^3 A}}$

We find that the ratio simplifies to $\phi ^3 = ({\frac {1 + \sqrt{5}}{2}})^3 = 2 + \sqrt{5}$

So $a = 2 , b = 5 , c = 1$ therefore $a + b + c = \boxed{8}$