induction - Show the Fibonacci numbers satisfy F(n) $ge$ $2 ^ {(n-1) / 2}$

Use induction to show that the Fibonacci numbers satisfy F(n) $\ge$ $(2 ^ {(n-1) / 2})$ for all n $\ge$ 3

My work thus far:

Base Case: F(3) $\ge$ $2 ^ {(3-1) / 2}$ => F(3) $\ge$ $2 ^ {1}$

Induction Hypothesis: Assume F(n) is true for all 3 < n < k

Inductive step: for (k + 1), F(k + 1) $\ge$ $2 ^ {(k + 1 - 1) / 2}$ =>
F(3) $\ge$ $2 ^ {k / 2}$

I'm not sure where to go from here.

Answer

You seem to be misunderstanding how a proof by induction works. Say you have a proposition $P(n)$ to be verified for all natural numbers $n=0,1,2,\dots$. The base case is to show that $P(0)$ is true, which is usually the easiest. The inductive hypothesis is that for some $k$, $P(n)$ is true for all $n=0,1,2,\dots,k$. You seem to get the idea until now. The inductive step is to show, assuming $P(0),P(1),\dots,P(k)$, that $P(k+1)$ is also true. This is what allows the domino reaction to occur: once $P(k+1)$ is true, so is $P(k+2)$, and so on for every natural number.

In this case, $F(k+1)\geq2^{(k+1-1)/2}$ is what you want to show. You do not a priori know it to be true. Thus, it makes no sense to start by assuming it, as you seem to have done. Instead, start with the assumptions that
$$ F(k-1)\geq 2^{(k-2)/2}\quad\text{and}\quad F(k)\geq 2^{(k-1)/2}.$$
Note that by definition, $F(k+1)=F(k)+F(k-1)$. Summing the above two inequalities,
$$F(k+1)=F(k)+F(k-1)\geq 2^{(k-2)/2}+2^{(k-1)/2}.$$

The big question, now, is whether it is true that $2^{(k-2)/2}+2^{(k-1)/2}$ is greater than or equal to $2^{k/2}$. It turns out that it is true. To show this is the inductive step that you have to make and the conclusion that would complete the proof. Can you finish now?

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Try completing the inductive step on your own first. If you need a hint, look below:

Note that $2^{(k-1)/2}>2^{(k-2)/2}$, so $2^{(k-1)/2}+2^{(k-2)/2}>2\cdot2^{(k-2)/2}=2^{(k-2)/2+1}=2^{k/2}$.