Let $m=\min(a_1,\ldots,a_n)$ and $M=\max(a_1,\ldots,a_n)$. Assume FTSOC that $M>m$. Then, there must exist an index $i$ such that $a_i=m$ and $\max(a_{i-1},a_{i+1})>m$, giving $b_i>2$. Similarly, there must exist an index $j$ such that $a_j=M$ and $\min(a_{j-1},a_{j+1})<M$, giving $b_j<2$. This means $a_i<a_j$ and $b_i>2>b_j$, a contradiction. Therefore, we have $M=m$, so $a_1=\cdots=a_n$.

From the given condition we get that $(b_{i+1}-b_i)(a_{i+1}-a_{i})\geq 0$ for $i=1,..,n$. Hence $\sum_{i=1}^n(b_{i+1}-b_i)(a_{i+1}-a_{i})\geq 0\Leftrightarrow $ $\Leftrightarrow\sum_{i=1}^nb_{i+1}a_{i+1}-\sum_{i=1}^nb_{i+1}a_i-\sum_{i=1}^nb_ia_{i+1}+\sum_{i=1}^na_ib_i\geq 0$ $\Leftrightarrow 2\sum_{i=1}^na_i\dfrac{a_{i-1}+a_{i+1}}{a_i}\geq \sum_{i=1}^na_i\dfrac{a_i+a_{i+2}}{a_{i+1}}+\sum_{i=1}^na_{i+1}\dfrac{a_{i-1}+a_{i+1}}{a_i}\Leftrightarrow$ $\Leftrightarrow 4\sum_{i=1}^n a_i\geq \sum_{i=1}^n \dfrac{a_i^2+a_ia_{i+2}}{a_{i+1}}+\sum_{i=1}^n\dfrac{a_{i+2}^2+a_ia_{i+2}}{a_{i+1}}\Leftrightarrow$ $\Leftrightarrow 4\sum_{i=1}^n a_i\geq \sum_{i=1}^n \dfrac{(a_i+a_{i+2})^2}{a_{i+1}}\overset{C-S}{\geq }\dfrac{4\left (\displaystyle \sum_{i=1}^n a_i \right )^2}{\displaystyle \sum_{i=1}^n a_i}=4\sum_{i=1}^n a_i$ so $\dfrac{a_i+a_{i+2}}{a_{i+1}}$ must be constant, i.e $b_1=b_2=..=b_n$. Now since $b_i\geq b_j$ and $b_j\geq b_i$ for all $i,j$ we get $a_i\geq a_j$ and $a_j\geq a_i$ for all $i,j$ and the result follows.

Let $b_k = \max(b_1,\dots, b_n)$. For any $i$, $b_k \ge b_i \implies a_k \ge a_i$, and hence $a_k \ge a_{k-1}$ and $a_k \ge a_{k+1}$. Hence, $b_k= \frac{a_{k-1} + a_{k+1}}{a_k} \le 2$. Also, $b_k\cdot n \ge \sum_{i=1}^{n} b_i = \sum_{i=1}^{n} \frac{a_{i-1}}{a_i} + \sum_{i=1}^{n} \frac{a_{i+1}}{a_i} \overset{\text{AM-GM}}{\ge} n + n = 2n \dots [\clubsuit]$. Hence $b_k \ge 2$. Hence, we must have $b_k=2$. Hence, all equalities must hold in $[\clubsuit]$ and hence $a_1=a_2=\dots=a_n$ and we are done!

for any $k\in[n]$, $\sum_{i =1}^n a_ib_i \ge \sum_{i =1}^n a_ib_{i+k}$, then sum by $k$, $2n\sum_{i =1}^na_i\ge \sum_{k=1}^n\sum_{i =1}^n a_ib_{i+k}=\sum_{i =1}^n a_i\sum_{k=1}^nb_k$, $\sum_{k=1}^nb_k\le 2n$, but it $\ge2n$, same as #4.

Same as #2.

lol Suppose that the $a_i$ are not all equal and pick some minimal $a_i$ such that $a_{i-1}$ is not minimal (this clearly exists). Then pick $a_j$ such that $a_j$ is maximal. We have $$a_i \leq a_j \implies \frac{a_{i-1}+a_{i+1}}{a_i} \leq \frac{a_{j-1}+a_{j+1}}{a_j} \implies \frac{a_{i-1}+a_i}{a_i}\leq \frac{a_j+a_j}{a_j} \implies \frac{a_{i-1}}{a_i} \leq 1,$$contradiction. $\blacksquare$

The given condition gives that $\{a_i\},\{b_i\}$ are similarly ordered , so Chebyshev gives $$\frac{\sum{a_i}}{n}\frac{\sum{b_i}}{n}\leq \frac{\sum{a_ib_i}}{n} =\frac{\sum{a_{i+1}+a_{i-1}}}{n}=2\frac{\sum{a_i}}{n} \implies \sum{b_i} \leq 2n $$But $$\sum{b_i}\geq \sum{2\frac{\sqrt{a_{i+1}a_{i-1}}}{a_i}}\geq 2n $$by AM GM .Thus equality holds everywhere and we are done

Basically the same as everyone else XD Let $i$ and $j$ be such that $a_i$ and $a_j$ are respectively the min and the max of the $a_k$'s. The we have $b_i=\frac{a_{i-1}+a_{i+1}}{a_i}\ge \frac{a_i+a_i}{a_i}\ge 2$. And $b_j=\frac{a_{j-1}+a_{j+1}}{a_j}\le \frac{a_j+a_j}{a_j}\le 2$. This implies that $b_i\ge b_j$, so in fact we have $a_i\ge a_j$, which means that the maximum value of the $a_k$'s is smaller than the smallest one, and from this it follows that $a_1=a_2=\ldots=a_n$. $\square$

Assume that there exists $i \in \{1, \dots, n\}$ such that $a_i$ is minimal. Then \[b_i = \frac{a_{i-1} + a_{i+1}}{a_i} \geq \frac{a_i + a_i}{a_i} = 2. \]From \begin{align*} a_{i} \leq a_{i+1} &\Longleftrightarrow b_{i} \leq b_{i+1} \\ &\Longleftrightarrow 2 \leq \frac{a_i + a_{i+2}}{a_{i+1}} \leq 1 + \frac{a_{i+2}}{a_{i+1}} \\ &\Longleftrightarrow a_{i+1} \leq a_{i+2} \end{align*}we inductively deduce that $ a_{i} \leq a_{i+1} \leq \cdots \leq a_{i-1} \leq {a_i}$, as desired.

Let $m=\min(a_1\cdots a_n)$ and $M=\max(a_1\cdots a_n).$ Then, $$b_{index(M)}=\frac{a_{index(M)-1}+a_{index(M)+1}}{M}\leq \frac{M+M}{M}=2=\frac{m+m}{m}\leq \frac{a_{index(m)-1}+a_{index(m)+1}}{m}=b_{index(m)}.$$Thus, since the condition implies $b_{index(M)}\geq b_{index(m)}$, equality must hold, $a_{index(M)-1}=a_{index(M)+1}=M$, so both neighbors of the maximum are also equal to the maximum. However, we can then repeat this process with one of the new maximum numbers, all the way until we get that every number in the circle is equal to the maximum, QED.

Let $a_M = \max(a_1,a_2,...,a_n)$, and $a_m = \min(a_1,a_2,...,a_n)$ then, $$b_M = \frac{a_{M-1}+ a_{M+1}}{a_M}$$$$b_m = \frac{a_{m-1}+ a_{m+1}}{a_m}$$Now, $a_M \geq a_{M-1}$ and $a_M \geq a_{M+1} \implies 2 \geq b_M $. Similarly l, $b_m \geq 2$. Hence $2 \geq b_M \geq b_m \geq 2 \implies b_M=b_m \implies a_M=a_m$. Hence $a_1 = a_2 = \dots = a_n$

For the sake of contradiction, assume $\min\{a_1, a_2, \dots, a_n\} \neq \max\{a_1, a_2, \dots, a_n \}.$ This implies there exists $a_i$ that is minimum such that $\max\{a_{i-1}, a_{i+1}\} > a_i.$ Similarly, we also know there exists $a_j$ that is maximal such that $\min\{a_{j-1}, a_{j+1}\} <a_j.$ Observe \[b_i = \frac{a_{i-1}+a_{i+1}}{a_i} > 2 \text{ and } b_j = \frac{a_{j-1}+a_{j+1}}{a_j} <2.\]But since $a_i<a_j,$ we know $2>b_i<b_j<2,$ absurd. The result follows.

Let $m=\min(a_i)$, and $M=\max(a_i)$. The key is the following observation. Lemma: If $a_i=m$ and $a_j=M$, then $a_{i-1}=a_{i+1}=m$ and $a_{j-1}=a_{j+1}=M$. Proof: Since $a_i=m\leq M=a_j$, we know that $b_i\leq b_j$. But then $$2=\frac{m+m}{m}\leq \frac{a_{i-1}+a_{i+1}}{a_i}\leq \frac{a_{j-1}+a_{j+1}}{a_j}\leq \frac{M+M}{M}=2,$$so equality must hold, and the lemma follows. $\square$ Now the problem is clear. Start from any $a_i=m$ and $a_j=M$, to eventually find that all values of the sequence are simultaneously equal to $m$ and $M$, i.e the sequence is constant.

Assume FTSOC that the $a_i$ are not all equal. Then, there exist an $i$ such that $a_i$ is the maximum of the sequence and a $j$ such that $a_j$ is the minimum of the sequence, and $a_i \neq a_j$ from our assumption. Then, first note that we have $b_i=\frac{a_{i-1}+a_{i+1}}{a_i} \leq \frac{a_j+a_j}{a_j}=2$. We also have $b_j=\frac{a_{j-1}+a_{j+1}}{a_j} \geq \frac{b_j+b_j}{b_j}=2$, so we can actually combine our two inequalities to obtain $2 \geq b_i \geq b_j \geq 2$, so in fact this is an equality and $b_i=b_j$, which implies $a_i=a_j$, contradiction. Thefore, all the $a_i$ are equal. $\blacksquare$