Inequality of the Fibonacci sequence and the golden ratio How can I prove that for each $n\in\Bbb Z^+$ $$\frac{f_{2n}}{f_{2n-1}}\leq\frac{1+\sqrt{5}}{2}$$ where each $f_i$ is a term of the Fibonacci sequence. 
Any help is really appreciated
 A: $$\frac{a^{2n}-b^{2n}}{a^{2n-1}-b^{2n-1}}-a=\frac{b^{2n-1}(a-b)}{a^{2n-1}-b^{2n-1}}$$
will be $<0$ for real $a-b>0, b<0$
Using Euler-Binet Formula,  $a=\dfrac{1+\sqrt5}2>0,b=\dfrac{1-\sqrt5}2<0\implies a-b>0$
So, $$\frac{a^{2n}-b^{2n}}{a^{2n-1}-b^{2n-1}}<a$$ for integer $n\ge0$
A: Use induction. For $n=1$, the statement is obviously true. For $n+1$, use these two facts:


*

*You already know that $\frac{f_{2n}}{f_{2n-1}}\leq \frac{1+\sqrt5}{2}$

*You know that $f_{2(n+1)} = f_{2n+2} = f_{2n+1} + f_{2n}$ and that $f_{2n+1} = f_{2n} + f_{2n-1}$.


Using this, you can prove the inductive step.
A: To me it seems reasonable to try to prove somewhat stronger claim by induction. (It happens quite often that trying to prove stronger statement might make inductive proof easier.)

For each $n$ the inequalities $$\frac{F_{2n}}{F_{2n-1}}\le\frac{1+\sqrt5}2 \qquad\text{and}\qquad \frac{F_{2n+1}}{F_{2n}}\ge\frac{1+\sqrt5}2$$ hold.

Inductive step should be relatively easy using the fact that the golden ratio $\varphi=\frac{1+\sqrt5}2$ fulfills the equation $\varphi^2-\varphi-1=0$. In particular, we have $\frac{\varphi+1}{\varphi}=\varphi$.
