# Prove that Fibonacci numbers are greater than a certain value

I was making good headway with a problem but I'm temporarily stuck,

$F_n$ is a certain number of a fibonacci sequence in this case.

I'm trying to prove: $F_n > (3/2)^{n-1}, n >= 6$

I've decided to do induction

Base case at 6 checks out

Inductive hypothesis $F_k > (3/2)^{k-1}, 6 <= k <= n$

Inductive step $F_{k+1} > (3/2)^{k}$

$F_{k+1} = F_{k} + F_{k-1}$ ...

Stuck here, because I can't attribute a value to $F_{k-1}$

Any help is appreciated! thank you!

• – lhf Apr 7 '17 at 0:26

Hint: Note that $$\left(\frac{3}{2}\right)^{n-1}+\left(\frac{3}{2}\right)^{n-2}>\left(\frac{3}{2}\right)^{n}\text{ for all }n$$ (to see this, multiply both sides by $(\frac{3}{2})^{-n}$ and simplify).

• Ah! very clever... almost didn't catch this until I started throwing equations at the wall to see what sticks – Howard P Apr 7 '17 at 0:48

$F_n = {\left( {1 + \sqrt{5} \over 2}\right)^n - \left({1 - \sqrt{5} \over 2}\right)^{-n} \over \sqrt{5}}$

where $\Phi = {1 + \sqrt{5} \over 2}$ is the Golden ratio.

Work from that.

Look at the difference:

$\left(\frac{3}{2}\right)^{n-1}-\frac{\left(\frac{1}{2} \left(1+\sqrt{5}\right)\right)^n-\left(\frac{1}{2} \left(1-\sqrt{5}\right)\right)^{-n}}{\sqrt{5}}$.

For $n=6$ it is positive (7.59375).

Then study the $n$-dependence of the difference... can it ever be negative?