TITLE: Show that $b\ge f_{n+1}$.
QUESTION [7 upvotes]: The Fibonacci sequence is defined as $f_1=f_2=1, f_{n+2}=f_{n+1}+f_{n}$. Assume that ${f_n \over f_{n-1}} \lt {a \over b} \lt {f_{n+1} \over f_n}$ ($a$ and $b$ are positive integers with no common prime factors). Show that $b\ge f_{n+1}$.

REPLY [2 votes]: Suppose $a,a',a''$ are non-negative integers and $b,b',b''$ are positive integers with $a''b'-b''a'=1$ and $\frac {a'}{b'}<\frac {a}{b}<\frac {a''}{b''}.$ Then $$0<ab'-a'b \quad \text { and }\quad 0<a''b-ab''.$$ Since these are all integers we have $$1\leq ab'-a'b \quad \text { and }\quad 1\leq a''b-ab''.$$ Multiplying the left inequality above by $b'',$ and multiplying the right inequality above by $b'$ we have $$b''\leq b''ab'-b''a'b \quad \text { and }\quad b'\leq b'a''b-b'a'b''.$$ Adding the above two inequalities we have  $$b''+b'\leq (b''ab'-b''a'b)+(b'a''b-b'ab'')=b(a''b'-b''a')=b.$$
Since $f_n^2-f_{n-1}f_{n+1}=(-1)^{n-1},$ if $(a',b')=(f_n,f_{n-1})$ and $(a'',b'')=(f_{n+1},f_n)$ and if $\frac {a''}{b''}>\frac {a'}{b'}$, then  $a''b'-b''a'=1.$