How to establish this inequality without using induction? Given the Fibonacci sequence $a_1 = 1$, $a_2 = 2$, $\ldots$, $a_{n+1} = a_n + a_{n-1} $ for $n \geq 2$, how to derive, without using induction, the inequality $$ a_n < (\frac{1+\sqrt{5}}{2})^n $$ for $n = 1, 2, 3, \ldots$? 
I know how to establish the above inequality using induction. 

The fibbonacci numbers have a closed form: $a_n = \dfrac{1}{\sqrt{5}}\left[\left(\dfrac{1+\sqrt{5}}{2}\right)^{n+1} - \left(\dfrac{1-\sqrt{5}}{2}\right)^{n+1}\right]$. 
Since $\left|\dfrac{1-\sqrt{5}}{2}\right| < 1$, we have $-1 < \left(\dfrac{1-\sqrt{5}}{2}\right)^{n+1} < 1$ for all $n \ge 1$. 
Can you figure out what to do from here?
