Consider the Fibonacci number $1,2,3,5,8,13,21,\cdots.$ It is well known that the limit of the ratio of Fibonacci numbers tends to the Golden Ratio $\phi$. Today, I want to show that $1.5$ is the smallest ratio between any two fibonacci numbers (achieved by $\frac{3}{2}$).
It is evident by inspection, but I was looking for a more formal proof of this.
I've tried using an induction argument, but it hasn't worked. Any ideas/suggestions would be greatly appreciated.
Hint: Consider the ratio of the ($n+1$)-th and $n$-th Fibonacci numbers: $$ \frac{a_{n+1}}{a_{n}}=\frac{a_{n}+a_{n-1}}{a_{n}}=1+\frac{a_{n-1}}{a_{n}}. $$ This quantity is at least as large as $3 / 2$ if only and if $2a_{n-1}\geq a_{n}$. Or, written in another way, $$ a_{n-1}+a_{n-1}\geq a_{n}. $$ Can you finish?