Fibonacci: prove the following theorem: define the Fibonacci sequence $\left\{ a_n\right\}_{n=0}^{\infty}$ by $a_0=a_1=1$ and for integers $k>1$, $a_k=a_{k-1}+a_{k-2}$. Then, for each integer $n$, $a_n\geq \left(\displaystyle\frac{3}{2}\right)^{n-2}$
Alright here is what I have so far.
(Basis Step)
$a_0 = 1 \geq 4/9 $
$a_1 = 1 \geq 2/3 $
(Inductive Hypothesis)
Fix $m\geq2$ and assume $a_n\geq \left(\displaystyle\frac{3}{2}\right)^{n-2}$ for all n greater than m.
Based on this definition $a_m=a_{m-1}+a_{m-2}$, we than have $a_{m-1} \geq \left(\displaystyle\frac{3}{2}\right)^{m-1-2}$ and $a_{m-2} \geq \left(\displaystyle\frac{3}{2}\right)^{m-2-2}$.
$a_m=a_{m-1}+a_{m-2} \geq \left(\displaystyle\frac{3}{2}\right)^{m-1-2} + \left(\displaystyle\frac{3}{2}\right)^{m-2-2}$.
Break
Alright that is what I have so far. I know that I have to use some algebraic manipulation to $\left(\displaystyle\frac{3}{2}\right)^{m-1-2} + \left(\displaystyle\frac{3}{2}\right)^{m-2-2}$ so that it looks like $\left(\displaystyle\frac{3}{2}\right)^{m-2}$.
However, I don't know how. My professor told me to factor out a $\left(\displaystyle\frac{3}{2}\right)^{m-2-2}$, but I don't know how it would look like.
This factors as $\left( \frac{3}2 \right)^{m-4} \left( 1+\frac{3}2 \right) = \left( \frac{3}2 \right)^{m-4} \left( \frac{5}2 \right) > \left( \frac{3}2 \right)^{m-2}$.