Let $b_k=\frac{(a_1+a_2)(a_2+a_3)\cdots(a_{k-1}+a_k)(a_k+a_1)}{a_1a_2...a_k}$ $2^k \leq b_k \leq 2^k+2022$ $\frac{b_{k+1}}{b_k}=\frac{(a_k+a_{k+1})(a_{k+1}+a_1)}{(a_k+a_1)a_{k+1}} = 2+\frac{(a_{k+1}-a_k)(a_{k+1}-a_1)}{a_{k+1}(a_k+a_1)} \geq 2$ So $b_{n+m} \geq 2^{m}b_n$ for every $n,m>0$ and so $b_n \leq \frac{2^{n+m}+2022}{2^{m}}=2^n+\frac{2022}{2^m}$ and for $m \to \infty$ we have $b_n=2^n$ But for $b_n=2^n$ we have that $\{a_n\}$ is constant.

Suppose ${a_n}$ is not constant. $\exists i \;\;s.t. \;\;a_i=a\neq a_{i+1}=b$ We can rewrite the condition in the following way. $$1+\frac{2022}{2^k}\ge \frac{a_1+a_2}{2\sqrt{a_1a_2}} \;\frac{a_2+a_3}{2\sqrt{a_2a_3}}\cdots \frac{a_k+a_1}{2\sqrt{a_ka_1}}$$Right side $\ge \frac{a+b}{2\sqrt{ab}}\;$ by AM-GM. ($k\ge i+1$) This is a constant value bigger than 1, but the left side converges to 1 when $k\to \infty$. Contradiction. Therefore ${a_n}$ has to be constant.

sketch: Take a really large N. Smooth the sequence into one where each term is the geometric mean of the term before it and the one after it. Thus it is geometric. In the sequence is non constant, then it is a geometric sequence with common ratio greater than 1 which can be shown to fail

考虑一个正数数列$\{a_n\}$满足一下两个条件 $1.$数列非严格递增 $2.$对任何正整数$k\geq3$ 下面式子成立 $$(a_1+a_2)(a_2+a_3)\cdots(a_{k-1}+a_k)(a_k+a_1)\leq (2^k+2022)a_1a_2\cdots a_k$$求证：${a_n}$是常数列