Show by induction that $F_n \geq 2^{0.5 \cdot n}$, for $n \geq 6$ I have the following problem:

Show by induction that $F_n \geq 2^{0.5 \cdot n}$, for $n \geq 6$

Where $F_n$ is the $nth$ Fibonacci number. 

Proof
Basis
$n = 6$. 
$F_6 = 8 \geq 2^{0.5 \cdot 6} = 2^{\frac{6}{2}} = 2^3 = 8$
Induction hypothesis
Assume $F_n \geq 2^{\frac{n}{2}}$, for some $n \geq 6$.
Inductive step
Lets shows that $F_{n+1} \geq 2^{\frac{n + 1}{2}}$. 
We know that 


*

*$F_{n + 1} = F_n + F_{n - 1}$

*$2^{\frac{n + 1}{2}} > 2^{\frac{n}{2}}$

*$2^{\frac{n + 1}{2}} = 2^{\frac{n}{2} + \frac{1}{2}} = 2^{\frac{n}{2}} \cdot 2^{\frac{1}{2}} \leq 2^{\frac{n}{2}} \cdot 2 = 2^{\frac{n}{2}} + 2^{\frac{n}{2}}$
Since we have assumed that $F_n \geq 2^{\frac{n}{2}}$, then $$F_n + F_{n - 1} = F_{n + 1} \geq 2^{\frac{n}{2}} +  F_{n - 1} \geq 2^{\frac{n}{2}} +  F_{n - 1} + 2^{\frac{n}{2}} - F_n = 2^{\frac{n}{2}} + 2^{\frac{n}{2}} +  F_{n - 1} - F_n$$
The last inequality is true because $2^{\frac{n}{2}} - F_n$ is negative or $0$, since $F_n \geq 2^{\frac{n}{2}}$.

I have tried a lot of things, but I cannot figure out how to proceed and how to conclude that indeed $F_{n + 1} \geq 2^{\frac{n}{2}} \cdot 2^{\frac{1}{2}}$. I feel really stupid after trying for a long time to do this alone and not managing to do it.

We have that $F_n>F_{n-1}$ then $$F_{n+1}=F_n+F_{n-1}>2F_{n-1}>2\cdot2^{(n-1)/2}=2^{(n+1)/2}$$
