Prove for all integers, by induction

Question

Define $\alpha = (1+ \sqrt{5})/2$. Define f(0) = 0, f(1) = 1, and f(n) = f(n-1) + f(n-2) for all $n\ge 2$.

Prove, for all integers $n\ge 3, f(n) \gt \alpha^{n-2}$

I understand the base case, but how do I solve the inductive case?

Answer 1

We need strong induction here

Let $f(n)>\alpha^{n-2}$ for $1\le r\le m+1$

So, $f(m+2)=f(m+1)+f(m)>\alpha^{m-1}+\alpha^{m-2}=\alpha^{m-2}(\alpha+1)=\alpha^{m-2}\cdot\alpha^2$