How to prove a Fibonacci inequality using Strong Induction? Using strong induction I am trying to prove that
$$F_n \geq \left(\frac{1+\sqrt{5}}{2}\right)^{n-2} \text{ for all } n \geq 2$$
for the Fibonacci Sequence defined by: $F_0 = 0$, $F_1 = 1$,  and $F_n = F_{n-1} + F_{n-2}$ for $n \geq 2$.
I know how to do strong induction for normal sequences, but not for inequalities (I have never liked inequalities).
If someone wouldn't mind pointing me in the right direction regarding what to do, I would be very grateful.

\begin{align}
F_n & = F_{n-1}+F_{n-2} \\[8pt]
& \ge \left(\frac{1+\sqrt{5}}{2}\right)^{n-3} + \left( \frac{1+\sqrt{5}}{2} \right)^{n-4} & & \text{(by the induction hypothesis)} \\[8pt]
& = \left(\frac{1+\sqrt{5}}{2}\right)^{n-4} \left( \frac{1+\sqrt{5}}{2} + 1 \right) & & \text{(Here we just pulled out a common factor.)} \\[8pt]
& = \left(\frac{1+\sqrt{5}}{2}\right)^{n-4} \left( \frac{1+\sqrt{5}}{2} \right)^2 & & \text{Where did this come from?  See below.} \\[8pt]
& = \left(\frac{1+\sqrt{5}}{2}\right)^{n-2}
\end{align}
We used the equality $\dfrac{1+\sqrt{5}}{2} + 1 = \left( \dfrac{1+\sqrt{5}}{2} \right)^2$.
To see this, just simplify both sides completely.  In both cases, it comes to $\dfrac{3+\sqrt 5} 2$.
