Fibonacci Number greater than Golden Section to Power less Two

Theorem
For all $n \in \N_{\ge 2}$:
 * $F_n \ge \phi^{n - 2}$

where:
 * $F_n$ is the $n$th Fibonacci number
 * $\phi$ is the golden section: $\phi = \dfrac {1 + \sqrt 5} 2$

Proof
The proof proceeds by induction.

For all $n \in \N_{\ge 2}$, let $\map P n$ be the proposition:
 * $F_n \ge \phi^{n - 2}$

Basis for the Induction
$\map P 2$ is true, as this just says:
 * $F_2 = 1 = \phi^0 = \phi^{2 - 2}$

It is also necessary to demonstrate $\map P 3$ is true:
 * $F_3 = 2 \ge \dfrac {1 + \sqrt 5} 2 = \phi = \phi^{3 - 1}$

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown that, if $\map P k$ is true, for all $1 \le k \le n$, then it logically follows that $\map P {k + 1}$ is true.

So this is the induction hypothesis:
 * $F_k \ge \phi^{k - 2}$

from which it is to be shown that:
 * $F_{k + 1} \ge \phi^{k - 1}$

Induction Step
This is the induction step:

So $\map P k \implies \map P {k + 1}$ and the result follows by the Second Principle of Mathematical Induction.

Therefore:
 * $\forall n \in \N_{\ge 2}: F_n \ge \phi^{n - 2}$

Also see

 * Fibonacci Number less than Golden Section to Power less One