Strong inductive proof for this inequality using the Fibonacci sequence. Problem
I need to determine for what natural numbers is $2n < F_n$, where $F_n$ is the $n^{th}$ Fibonacci number determined by $F_0 = 0$, $F_1 = 1$ and $F_n = F_{n-1}+F_{n-2}$.  I then need to prove my findings through strong induction.
What I found
I found that the inequality is true for all $n >= 8$.
My attempt at proving by induction
Basis: $2(8) < F_8$ = TRUE
Assume: $2(k) < F_k$
Show: $2(k) < F_k$ implies $2(k+1) < F_{k+1}$ 
$2(k+1) = 2k + 2 < F_k + F_{k-1} = F_{k+1}$
Thus
$2(k+1) < F_{k+1}$ 
Logic:
$2k < F_k$ by induction hypothesis
$2 < F_{k-1}$ because $F_{k-1}$ is at least $13$ when $k>=8$
$F_{k+1}$ is $F_k + F_{k-1}$.
Is my proof correct?  Is this considered strong induction?
 A: Your proof is great. Here's how you would explicitly use strong induction. Note that you have already proved the base case for when $n=8$.
Induction Hypothesis: Assume that $F_n>2n$ holds true for all $n\in\{8,...,k\}$, where $k\ge8$.
It remains to prove the inequality true for $n=k+1$. Observe that:
$$ \begin{align*}
F_{k+1} &= F_k + F_{k-1} \\
&> 2k + 2(k-1) & \text{by the induction hypothesis} \\
&\ge 2k + 2(8-1) & \text{since } k \ge8 \\
&= 2k+14 \\
&> 2k+2 \\
&= 2(k+1) \\
\end{align*} $$
as desired. This completes the induction.
A: IIRC, strong induction is when the induction depends on
more than just the preceding value.
In this case, 
you use the hypothesis for $k$ but 
not for any earlier values.
Instead, you use a much weaker result
($F_{k-1} > 2$) for the earlier value.
So, I would not call this strong induction.
If you use the hypothesis
($F_n > 2n$) for $both$
$k$ and $k-1$,
 the induction works because
$F_k > 2k$ and $F_{k-1} > 2(k-1)$
together imply
$F_{k+1} = F_k+F_{k-1}
> 2k + 2(k-1)
= 4k-2
> 2(k+1)
$
when
$k \ge 3$.
Note that the induction step works when
$k \ge 3$
but the induction hypothesis
is true only when
$k \ge 8$.
So the first case where you can
do the induction is
$k = 9$,
because you use the truth for
$k=8$ and $k=9$ to prove it for
$k=10$.
I would call this moderate induction,
since it depends on 
the previous two cases being true.
A: We will change the indexing slightly to make it look like archetypal strong induction. 
We show that if the result holds for all $i$ such that $8\le i \lt k$ then it holds at $k$. 
Because it holds at for all such $i$, it holds in particular at $k-1$. Now argue as you did (with minor index shift) that since $F_{k-1} \gt 2(k-1)$ and $2\lt F_{k-2}$, we have $2k=2(k-1)+2\lt F_{k-1}+F_{k-2}=F_k$.
Of course we have not used the full strength of the induction hypothesis, but we have written out things in strong induction style. 
A: $F_n > 2n$ if $n > 8$. 
Proof:
Consider $F_n = F_{n-1} + F_{n-2}$
Thus,
$$F_{n+1} = F_n + F_{n-1}$$ 
The idea is to show:
$$F_{n+1} > 2n + 2$$
By strong induction, for $n > 8$ since $P(8)$ is true,
$$F_n > 2n$$
$$F_{n-1} > 2n + 2$$
$$F_{n+1} = F_{n} + F_{n-1} > 4n + 2$$
Lemma: $4n + 2 > 2n + 2$
Proof:
$$4n + 2 > 2n + 2$$
$$4n > 2n$$
$$4 > 2$$ 
Hence,
$$F_{n+1} > 2n + 2$$
