We show that the sequence is bounded above. In fact, this finishes because if $a_1,a_2,\ldots$ satisfies the given constraint then so does $1/a_1,1/a_2,\ldots$, and we can apply the same argument to that sequence. First note that $\frac{a+b}{1+ab} \geq 1$ if and only if $a-1,b-1$ have different sign, so for any fixed value of $i+j$, $(a_i-1)(a_j-1)$ should have fixed sign (positive, negative, or zero). Clearly if $a_i=1$ then $a_j=1$ for all $j>i$ as well, else take $(i,2j-i)$ and $(j,j)$, and then the desired claim is true. Thus assume that no terms in the sequence are $1$. Then if $a_i<1$ then $a_{i+2}<1$ is forced by comparing $(i,i+2)$ with $(i+1,i+1)$, and by induction eventually either every term is at most $1$ or exactly other term is at most $1$. We only care about the latter scenario, in which case we care only about the subsequence $a_k,a_{k+2},\ldots$ of terms greater than $1$, which "behaves" the same way as a complete sequence in itself would. Thus assume that every term is greater than $1$. Let $a_i=\coth x_i$ for each $i$, so each $x_i$ is a positive, and it suffices to show $x_i$ is bounded below by a positive real. The condition (after reciprocating both sides) then becomes $\coth(x_k+x_n)=\coth(x_m+x_l) \implies x_k+x_n=x_m+x_l \implies 2x_n=x_{n-1}+x_{n+1}$. This is then a linear recurrence whose solutions (over $\mathbb{R}$) are $x_n=an+b$. But if $a<0$ then we eventually have $x_n<0$ which is forbidden, so $a \geq 0$ and the $x_i$ are bounded below by $x_1>0$. $\blacksquare$