A Fibonacci conjecture: $\log_{F_{n+1}}{F_n}<\log_{F_{n+2}}{F_{n+1}}$ Given the Fibonacci sequence $F_n$, that is $$F_1=F_2=1,F_{n+2}=F_{n+1}+F_n.$$
I find that
$$
\log_21&lt;\log_32&lt;\log_53&lt;\log_85&lt;\cdots,
$$
i.e. we have $$
\log_{F_{n+1}}{F_n}&lt;\log_{F_{n+2}}{F_{n+1}}, \quad \text{for} \; n\geqslant1\tag{1}
$$
If $n=2k$, $(1)$ is easy to get, but if $n=2k+1$, I have no idea to prove it.
What's more, if we let $F_1=a,F_2=b$ be two positive integers, then $(1)$ still exstis when $n$ is greater than some integer $m$.
I do not konw whether my conjecture is true.
For even $n$ you can use:
$$F_nF_{n+2}&lt;F_{n+1}^2$$
$$\implies \log(F_nF_{n+2})&lt;\log(F_{n+1}^2)$$
$$\implies \log(F_n)+\log(F_{n+2})&lt;2\log(F_{n+1})$$
$$\implies \frac{\log(F_n)+\log(F_{n+2})}{2}&lt;\log(F_{n+1})$$
By the AM-GM:
$$\sqrt{\log(F_n)\log(F_{n+2})}&lt;\frac{\log(F_n)+\log(F_{n+2})}{2}&lt; \log(F_{n+1})$$
as required.
Conjecture:$ \log_{F_{n+1} } F_n &lt; \log_{F_{n+2} }F_{n+1}.$
This is equivalent to:
$$\dfrac{\log F_{n}}{\log F_{n+1}} &lt; \dfrac{\log    F_{n+1}}{\log F_{n+2}}$$
Multiply both sides by $\log F_{n+1}\log F_{n+2}$ and you'll get  $\log(F_n )\log (F_{n+2}) &lt; (\log (F_{n+1}))^2$ ($\log$ here is the natural logarithm).
One can use the Cassini identity ($F_{n+1}^2 = F_nF_{n+2} \pm 1$) to prove your conjecture, but first,
$$\log(F_n)\log(F_{n+2})&lt; (\log(F_{n+1}))^2 = (\frac{1}2\log(F_{n+1}^2))^2 = \frac{1}{4}(\log(F_{n+1})^2)^2$$
Case 1: when $F_{n+1}^2 = F_nF_{n+2} - 1$
$$4\log(F_n)\log(F_{n+2})&lt; (\log(F_{n+1})^2)^2 = \log(F_nF_{n+2} - 1)^2 &lt; (\log F_n + \log F_{n+2})^2.$$
Case 2: when $F_{n+1}^2 = F_nF_{n+2} + 1$
$$ 4\log(F_n) \log(F_{n+2})&lt; (\log(F_{n+1})^2)^2 = (\log(F_n F_{n+2} + 1))^2 &lt; (\log(F_n F_{n+2}) + 1)^2 = (\log F_n + \log F_{n+2})^2 + 2\log(F_n F_{n+2}) + 1.$$