Fibonacci numbers prove by induction that $F(n)>2n$ for $ n>7$ So I am having trouble proving that $F(n)>2n$ for all $n>7$. 
for base case I took n=8, so obviously $f(8)=21> 2*8= 16$
Assuming$ f(n)>2n$
$f(n+1)>2(n+1)$
$f(n)+f(n-1)>2n+2$
$f(n)> 2n$ according to the induction hypothesis and since $n>7$ ,$f(n-1)$ should be at least $f(7)$ and therefore more than 2, however I'm not sure if this is the correct way to prove this by induction on n... 
 A: Assume $$f_n>2n$$ and $$f_{n+1}>2(n+1)=2n+2$$ for $n\ge 8$.Then you have
$$f_{n+2}=f_n+f_{n+1}>2n+2n+2>2n+4=2(n+2)$$
together with the start $f_8>16$ and $f_9>18$ this completes the proof
A: Another way :
It is easier to prove : 
$$\begin{array}{l}
{F_n} = \frac{{{\varphi ^n} - {\psi ^n}}}{{\varphi  - \psi }} = \frac{{{\varphi ^n} - {\psi ^n}}}{{\sqrt 5 }} > \frac{{{\varphi ^n}}}{{\sqrt 5 }} = \frac{{{{(\frac{{1 + \sqrt 5 }}{2})}^n}}}{{\sqrt 5 }}\\
\varphi  = \frac{{1 + \sqrt 5 }}{2} \approx 1.61803{\mkern 1mu} 39887 \cdots 
\end{array}$$
so prove that $$\forall n>7 :F_n>\frac{{{{(\frac{{1 + \sqrt 5 }}{2})}^n}}}{{\sqrt 5 }}>2n$$ so 
$$p(8) \space \frac{{{{(\frac{{1 + \sqrt 5 }}{2})}^8}}}{{\sqrt 5 }}>2\times 8 \\20.57>16 \checkmark $$
$$p(n) :\space \frac{{{{(\frac{{1 + \sqrt 5 }}{2})}^n}}}{{\sqrt 5 }}>2n\\p(n+1): \space  \frac{{{{(\frac{{1 + \sqrt 5 }}{2})}^{n+1}}}}{{\sqrt 5 }}>2(n+1)$$ multiply $p(n) $ by $\frac{{1 + \sqrt 5 }}{2}$
$$\space \frac{{{{(\frac{{1 + \sqrt 5 }}{2})}^{n+1}}}}{{\sqrt 5 }}>2n(\frac{{1 + \sqrt 5 }}{2})>2n(1.5)=\underbrace{3n >2n+2}_{\forall n>7} \checkmark$$
A: Consider the following proposition (P):
For all $n\in \Bbb N$ if $F_{n+8} > 2(n+8)$ and $F_{n+7} > 2(n+7)$ then 
$F_{n+9} > 2(n+9)$.  
Let us prove $P$ by induction: The basis is easy:  For $n=1$ this becomes
$F_{10} > 2(10)$ or $55> 20$.
Now assume (P) holds for all $n<k$.  Then 
$$
F_k = F_{k-1} + F_{k-2} > 2(k+8) +2(k+7) > 4k+30 > 2k+18 = 2(k+9)
$$
so (P) holds for $n=k$ as well; this establishes induction.
Thus (P) holds for all $n \in \Bbb N$, which in turns says that  $\forall n\geq 9 : F_n > 2n$.
The proof of the original statement is then completed by noting that the only omitted cases is $n=8$ and $F_8 = 21 > 16$.
