How do I prove $\frac{f_{2n+1}}{f_{2n}}>\frac{f_{2n}}{f_{2n-1}}$? How do I prove this inequality with sequential Fibonacci numbers: $\frac{f_{2n+1}}{f_{2n}}>\frac{f_{2n}}{f_{2n-1}}$ (i.e. $\frac{5}{3}>\frac{3}{2}$)? I know these formulas alternately overestimate and underestimate the golden ratio as $n$ gets large.
 A: $$\begin{align*}
f_{2n+1}f_{2n-1} - f_{2n}^2
&= \det\pmatrix{f_{2n+1}&f_{2n}\\f_{2n}&f_{2n-1}}\\
&= \det\pmatrix{f_{2n+1}-f_{2n}&f_{2n}\\f_{2n}-f_{2n-1}&f_{2n-1}}&&C_1 \to C_1-C_2\\
&= \det\pmatrix{f_{2n-1}&f_{2n}\\f_{2n-2}&f_{2n-1}}\\
&= \det\pmatrix{f_{2n-1}&f_{2n}-f_{2n-1}\\f_{2n-2}&f_{2n-1}-f_{2n-2}} && C_2 \to C_2-C_1\\
&= \det\pmatrix{f_{2n-1}&f_{2n-2}\\f_{2n-2}&f_{2n-3}}\\
&= \vdots\\
&= \det\pmatrix{f_{3}&f_{2}\\f_{2}&f_{1}}\\
&= \det\pmatrix{2&1\\1&1}\\
&= 1\\
&> 0\\
\frac{f_{2n+1}}{f_{2n}}&> \frac{f_{2n}}{f_{2n-1}}
\end{align*}$$
(for $n \ge 1$, so that $f_{2n-1}$ and $f_{2n}$ are both positive)
A: The characteristic equation $^2−−1=0$ has two roots $\varphi=\frac{1+\sqrt 5}{2}, \psi=\frac{1-\sqrt 5}{2}$. Vieta’s formulas give $\varphi+\psi=1, \varphi \psi = -1$. Then we can rewrite $f_{n+1} = f_n + f_{n-1}$ as
$$f_{n+1}−\varphi f_=\psi (f_−\varphi f_{-1}).$$
Note that $\psi<0$. Then
$$f_{+1}−\varphi f_=\psi^ (f_1− \varphi f_0 )=\psi^ \begin{cases} >0 & \text{ if } n \text{ is even;} \\ <0 & \text{ if } n \text{ is odd.} \end{cases}$$
Hence
$$
\frac{f_{2n+1}}{f_{2n}}>\varphi >\frac{f_{2n}}{f_{2n-1}}
$$
A: The base case of OP's equality is, for $n=1$,
$$\frac21>\frac11.$$
Assume that OP's inequality holds for some $n = k$, $k \ge 1$.
$$\frac{f_{2k+1}}{f_{2k}} > \frac{f_{2k}}{f_{2k-1}}$$
Then for $n=k+1$, using the mediant inequality twice,
$$\begin{array}{rcrclcl}
\dfrac{f_{2k+1}}{f_{2k}} &>&&&&& \dfrac{f_{2k}}{f_{2k-1}}\\
\dfrac{f_{2k+1}}{f_{2k}} &>&&& \dfrac{f_{2k+1}+ f_{2k}}{f_{2k}+f_{2k-1}} &> &\dfrac{f_{2k}}{f_{2k-1}}\\
\dfrac{f_{2k+1}}{f_{2k}} &>&&& \dfrac{f_{2k+2}}{f_{2k+1}}\\
\dfrac{f_{2k+1}}{f_{2k}} &>& \dfrac{f_{2k+1}+f_{2k+2}}{f_{2k}+f_{2k+1}} &>& \dfrac{f_{2k+2}}{f_{2k+1}}\\
&&\dfrac{f_{2k+3}}{f_{2k+2}} &>& \dfrac{f_{2k+2}}{f_{2k+1}}
\end{array}$$
(The intermediate step also proves the other half of the inequality pair:
$$\frac{f_{2n+2}}{f_{2n+1}} < \frac{f_{2n+1}}{f_{2n}})$$
