Given the Fibonacci sequence $F_n$, that is $$F_1=F_2=1,F_{n+2}=F_{n+1}+F_n.$$ I find that $$ \log_21<\log_32<\log_53<\log_85<\cdots, $$ i.e. we have $$ \log_{F_{n+1}}{F_n}<\log_{F_{n+2}}{F_{n+1}}, \quad \text{for} \; n\geqslant1\tag{1} $$ 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.
Conjecture: $ \log_{F_{n+1} } F_n < \log_{F_{n+2} }F_{n+1}.$
This is equivalent to: $$\dfrac{\log F_{n}}{\log F_{n+1}} < \dfrac{\log F_{n+1}}{\log F_{n+2}}$$ Multiply both sides by $\log F_{n+1}\log F_{n+2}$ and you'll get $\log(F_n )\log (F_{n+2}) < (\log (F_{n+1}))^2$ ($\log$ 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 let's establish an inequality equivalent to the above one, $$\log(F_n)\log(F_{n+2})< (\log(F_{n+1}))^2 = (\frac{1}2\log(F_{n+1}^2))^2 = \frac{1}{4}(\log(F_{n+1})^2)^2$$
Therefore it suffices to prove that $\log(F_{n+1}^2)^2 > 4\log F_n \log F_{n+2}$
Case 1: when $F_{n+1}^2 = F_nF_{n+2} - 1$ $$(\log(F_{n+2}/F_{n}))^2 > (\log 2)^2 > \dfrac{2 \log F_n}{F_n} + \dfrac{2\log 3F_n}{F_n} - \dfrac{1}{F_n^2}$$ $$\geq \dfrac{2 \log F_n + 2\log (2F_n + F_{n-1})}{F_n}- \dfrac{1}{F_n^2} = \dfrac{2 \log F_n + 2\log (F_{n+2})}{F_n}- \dfrac{1}{F_n^2}$$ $$\implies (\log F_{n+2} - \log F_n)^2 > \dfrac{2 \log F_nF_{n+2}}{F_n} - \dfrac{1}{F_n^2}$$ for a sufficiently large $n$($n \geq 8$ for the Fibonacci sequence starting with $1$ and $2$). Now add $4\log F_n \log F_{n+2}$ to both sides: $$(\log F_{n+2} + \log F_n)^2 > \dfrac{2 \log F_nF_{n+2}}{F_n} - \dfrac{1}{F_n^2} + 4\log F_n \log F_{n+2} $$ $$\implies (\log F_{n+2} + \log F_n)^2 - \dfrac{2 \log F_nF_{n+2}}{F_n} + \dfrac{1}{F_n^2} > 4\log F_n \log F_{n+2}$$ $$\implies (\log F_{n} + \log F_{n+2} - \dfrac{1}{F_n})^2 > 4\log F_n \log F_{n+2}$$ $$\implies (\log F_{n} + \log (F_{n+2} - \dfrac{1}{F_n}))^2 = (\log(F_n\cdot (F_{n+2} - \dfrac{1}{F_n})))^2 = \log(F_{n+1}^2)^2 > 4\log F_n \log F_{n+2} $$
Case 2: when $F_{n+1}^2 = F_nF_{n+2} + 1$ $$(\log F_{n+2} - \log F_n)^2 > 0 $$ $$\implies (\log F_{n+2} +\log F_n)^2 > 4\log F_n \log F_{n+2}$$ $$\implies (\log F_{n+1}^2)^2 = (\log F_{n+2}F_n + 1)^2 > (\log F_{n+2}F_n)^2 > 4\log F_n \log F_{n+2}$$
Verifying the inequality for $n < 8$ manually completes the proof using Cassini's identity.
For even $n$ you can use:
$$F_nF_{n+2}<F_{n+1}^2$$
$$\implies \log(F_nF_{n+2})<\log(F_{n+1}^2)$$
$$\implies \log(F_n)+\log(F_{n+2})<2\log(F_{n+1})$$
$$\implies \frac{\log(F_n)+\log(F_{n+2})}{2}<\log(F_{n+1})$$
By the AM-GM:
$$\sqrt{\log(F_n)\log(F_{n+2})}<\frac{\log(F_n)+\log(F_{n+2})}{2}< \log(F_{n+1})$$
as required.
In general, starting with any two numbers, you get
$$F_n=A\phi^n+B(-\phi)^{-n}$$
where $\phi=(1+\sqrt5)/2$ is the golden ratio. The second term soon becomes very small. So
$$\ln F_n\ln F_n-\ln F_{n-1}\ln F_{n+1}\approx \\
[n\ln\phi+\ln A]^2-[(n-1)\ln\phi+\ln A][(n+1)\ln\phi+\ln A]\\
\approx (\ln\phi)^2$$
Conjecture:$ \log_{F_{n+1} } F_n < \log_{F_{n+2} }F_{n+1}.$
This is equivalent to: $$\dfrac{\log F_{n}}{\log F_{n+1}} < \dfrac{\log F_{n+1}}{\log F_{n+2}}$$ Multiply both sides by $\log F_{n+1}\log F_{n+2}$ and you'll get $\log(F_n )\log (F_{n+2}) < (\log (F_{n+1}))^2$ ($\log$ 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, $$\log(F_n)\log(F_{n+2})< (\log(F_{n+1}))^2 = (\frac{1}2\log(F_{n+1}^2))^2 = \frac{1}{4}(\log(F_{n+1})^2)^2$$
Case 1: when $F_{n+1}^2 = F_nF_{n+2} - 1$ $$4\log(F_n)\log(F_{n+2})< (\log(F_{n+1})^2)^2 = \log(F_nF_{n+2} - 1)^2 < (\log F_n + \log F_{n+2})^2.$$
Case 2: when $F_{n+1}^2 = F_nF_{n+2} + 1$
$$ 4\log(F_n) \log(F_{n+2})< (\log(F_{n+1})^2)^2 = (\log(F_n F_{n+2} + 1))^2 < (\log(F_n F_{n+2}) + 1)^2 = (\log F_n + \log F_{n+2})^2 + 2\log(F_n F_{n+2}) + 1.$$
