If $\left\{a_n\right\}$ is Fibonacci Number, prove that the sequence$\left\{\frac{\ln a_n}{\ln a_{n+1}}\right\}$ increase 
If $a_0=1,a_1=1,a_{n+2}=a_{n+1}+a_{n}$, prove the sequence$\left\{\frac{\ln a_n}{\ln a_{n+1}}\right\},\,n\geq 1$ increase.

When $n$ is even this inequality is easy proved by Am-Gm.
If $n$ is even,we have $$\frac{\ln a_n}{\ln a_{n+1}}<\frac{\ln a_{n+1}}{\ln a_{n+2}}\Leftrightarrow \ln a_n\cdot\ln a_{n+2}<(\ln a_{n+1})^2$$
By Am-Gm, and $a_{n-1}a_{n+1}-a_n^2=(-1)^n$, we have$$LHS<\left(\frac{\ln a_n+\ln a_{n+2}}{2}\right)^2=\left(\frac{\ln(a_n\cdot a_{n+2})}{2}\right)^2=\left(\frac{\ln(a_{n+1}^2-1)}{2}\right)^2<RHS.$$
But when n is odd, the inequality is a little difficult, or I didn't use a right method.
Supplement:
Here we prove that $$a_{n-1}a_{n+1}-a_n^2=(-1)^n.$$
It is not hard to prove that if $$A=\begin{pmatrix}
0&1\\
1&1
\end{pmatrix}$$then$$A^n=\begin{pmatrix}
a_{n-1}&a_n\\
a_n&a_{n+1}
\end{pmatrix}$$by induction. So $$a_{n-1}a_{n+_1}-a_n^2=\det\begin{vmatrix}
0&1\\
1&1
\end{vmatrix}^n=\left(\det\begin{vmatrix}
0&1\\
1&1
\end{vmatrix}\right)^n=(-1)^n$$

There is a primary proof by a student in Fudan University(Not me).
First, it is easy to prove that $\left\{\frac{a_{n+1}}{a_{n}}\right\}$ strictly increase.
Second, we can also prove $$0\leq\ln(1+x)\leq x,\quad(\forall\,x\geq 0).$$
And now we begin the proof.
Consider $\forall\,n\geq 3$, we have $$\frac{a_{n+1}}{a_{n-1}}=\frac{a_{n}}{a_{n-1}}\cdot\frac{a_{n+1}}{a_n}\geq\frac{3}{2}\cdot\frac{3}{2}=\frac{9}{4}.$$
So
$$
\ln a_{n+1}-\ln a_{n-1}\geq\ln\frac{9}{4}=2\ln\frac{3}{2}>\frac{4}{5}\Rightarrow(\ln a_{n+1}-\ln a_{n-1})^2>\frac{16}{25}
$$
By this inequality when $n\geq 5$ we can get
$$
\begin{align*}
\ln a_{n+1}\cdot\ln a_{n-1}&=\frac{1}{4}(\ln a_{n+1}+\ln a_{n-1})^2-\frac{1}{4}
(\ln a_{n+1}-\ln a_{n-1})^2\\
&< \frac{1}{4}(\ln a_{n+1}+\ln a_{n-1})^2-\frac{4}{25}\\
&=\frac{1}{4}(\ln(a_{n+1}\cdot a_{n-1}))^2-\frac{4}{25}\\
&=\frac{1}{4}(\ln(a_{n}^2+(-1)^{n-1}))^2-\frac{4}{25}\\
&=\frac{1}{4}(\ln(a_{n}^2+(-1)^{n-1}))^2-\frac{1}{4}(\ln a_n^2)^2+(\ln a_n)^2-\frac{4}{25}\\
&\leq\frac{1}{4}(\ln(a_n^2+1))^2-\frac{1}{4}(\ln a_n^2)^2+(\ln a_n)^2-\frac{4}{25}\\
&=\frac{1}{4}(\ln(a_n^2+1)-\ln a_n^2)(\ln(a_n^2+1)+\ln a_n^2)+(\ln a_n)^2-\frac{4}{25}
\\
&\leq\frac{1}{4}\left(\ln(a_n^2+1)-\ln a_n^2\right)\cdot2\ln((a_n+1)^2)+(\ln a_n)^2-\frac{4}{25}\\
&=\ln(1+\frac{1}{a_n^2})\ln(1+a_n)+(\ln a_n)^2-\frac{4}{25}\\
&\leq \frac{1}{a_n^2}\cdot a_n-\frac{4}{25}+(\ln a_n)^2\\
&\leq\frac{1}{8}-\frac{4}{25}+(\ln a_n)^2\\
&<(\ln a_n)^2
\end{align*}
$$
In a word, it is $\ln a_{n-1}\cdot\ln a_{n+1}<(\ln a_n)^2$, which is equal to$$\frac{\ln a_{n-1}}{\ln a_n}<\frac{\ln a_n}{\ln a_{n+1}},\quad n\geq 5.$$
For the rest situation ,we just need to prove it by caculating it.
