Fibonacci, proof with induction

Hello, I think I am getting stuck; I would really appreciate any help or advice!

Prove by induction that $\displaystyle {Fib}_n \geq 2^{n/2}$ for $\displaystyle n \geq 6$

To start , prove the base case of $\displaystyle n = 6$:

$\displaystyle {Fib}_6 = 8, 2^{6/2} = 8$ so $\displaystyle {Fib}_6 \geq 2^{6/2}$

Now assume $\displaystyle {Fib}_n \geq 2^{n/2}$ is true.

We can multiply both sides by $\displaystyle 2^{1/2}$ to get the $\displaystyle n+1$ term on the right side.

$\displaystyle 2^{1/2}{Fib}_n \geq 2^{(n+1)/2}$

This is where I think I'm not sure about what to do. What I had been thinking is: now try to show that $\displaystyle {Fib}_{n+1} \geq 2^{1/2}{Fib}_n$

so that we can conclude with $\displaystyle {Fib}_n \geq 2^{n/2} \Longrightarrow {Fib}_{n+1} \geq 2^{(n+1)/2}$.

I had been trying: It's true that $\displaystyle {Fib}_{6+1} \geq 2^{1/2}{Fib}_6$. If we add $\displaystyle {Fib}_n$ to both sides, we can get the n+2 term on the left. But then I wasn't sure if there is another step..? Or if I'm on the right track at all?

THANKS! for any help!!