Fibonacci Number Proof How can I prove this statement? Would I use induction?
"Given $n \geq 11$, show that $a_n > (3/2)^{n}$. $a_n$ is the $n$th Fibonacci number."
 A: Yes, induction is the way to go. Assume the result is true for two consecutive integers $n$ and $n+1$ and then deduce that it must be true for $n+2$. The rest should be easy, after you find 2 consecutive values for which it is definitely true.
To explain a bit more:
Assume the result is true for $n$ and for $n+1$, i.e. assume we have $a_n > (3/2)^n$ and $a_{n+1} > (3/2)^{n+1}$.
Adding these two, we get $a_{n+2} = a_{n+1} + a_n > (3/2)^{n+1} + (3/2)^n = (3/2)^n(3/2 + 1) = (3/2)^n(5/2) > (3/2)^{n+2}$
at the last step we use the fact that $5/2 > 9/4 = (3/2)^2$
Now we know that if the result is true for $n$ and $n+1$, then it follows that it is true for $n+1$ and $n+2$.
A: Hint $\ $ The second order recurrence for $\rm\:f(n)\:$ yields one for $\rm\:f(n)-c^n,\:$ namely, more generally,
$$\begin{eqnarray}\rm  &&\rm f(n\!+\!2) &=&\rm\ a\ f(n\!+\!1)\ +\ b\ f(n)\\
\Rightarrow\ &&\rm f(n\!+\!2)-c^{n+2} &=&\rm\ a\,(f(n\!+\!1)-c^{n+1}\!)\ +\ b\,(f(n)-c^n)\ -\ c^n(\color{#C00}{c^2 - a\,c -b})\end{eqnarray}$$
So we can inductively infer positivity of the LHS from positivity of the $3\,$ summands on the RHS, which follows if $\rm\:a,b,c > 0\:$ and $\rm\:\color{#C00}{f(c)} < 0\:$ for the characteristic polynomial $\rm\:\color{#C00}{f(x)\, =\, x^2 - a\,x - b}.$  
In your case $\rm\:a,b,c\, =\, 1,1,3/2\, >\, 0,\:$ and $\rm\:\color{#C00}{f(c)} = (3/2)^2\!-3/2-1 =\, \color{#C00}{-1/4} < 0,\:$ so it succeeds.
