I need to show that $F_{n+2} \geq 2F_{n}$ for all $n \geq 1$.
For n=1, $F_3=2=2(F_1)=2$.Checks out.
Now for the inductive hypothesis, assume for some n that $F_{n+2} \geq F_{n}$. We need to show that $F_{n+3} \geq 2F_{n+1}$.
Here is where I run into trouble.
$F_{n+3}=F_{n+2}+F_{n+1}$, but it seems clear to me that $F_{n+2}>F_{n+1}$, giving the desired conclusion.
How can I use the inductive hypothesis here? it feels like I really don't need it.
Yes, it's kind of a silly problem as it is so obviously true ... induction is total overkill .... but you are asked to use induction, so induction it will be.
Here is the thing to do: since the Fibonacci sequence is recursively defined on two earlier entries, you need a base of two cases, and for the inductive hypothesis for the step, assume that the statement holds for both $n+1$ and $n+2$
In other words, first show that $F_3 \ge 2F_1$ and that $F_4 \ge 2F_2$
Then, assume that $F_{n+1} \ge 2F_{n-1}$ and $F_{n+2} \ge 2F_n$, and use that to show that $F_{n+3} \ge 2F_{n+1}$