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... 

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 $\frac{{1 + \sqrt 5 }}{2}$
$$\space \frac{{{{(\frac{{1 + \sqrt 5 }}{2})}^{n+1}}}}{{\sqrt 5 }}>2n(\frac{{1 + \sqrt 5 }}{2})>2n(1.5)=\underbrace{3n >2n+2}_{\forall n>7} \checkmark$$
