I am having trouble finding the lub and glb for the set
$$A_n = \Big\{ \sum_{k=1}^{n} \frac{a_k}{a_k + a_{k+1} + a_{k+2}} : a_1, a_2, ..., a_n \in \mathbb{R}_{+}, a_{n+1} = a_1, a_{n+2} = a_2 \Big\}$$
where $ n \geq 3$ is a fixed integer and $\mathbb{R}_+$ is the set of positive reals. The case when $n = 3$ is trivial, and I have conjectured that for all such $n$, $glb A_n = 1, lub A_n = n-2$, for, if we try to make each fraction as close to cero as we can by letting $a_2$ be much greater than $a_1$, $a_3$ be much greater than $a_2$, and so on, when we arrive to $a_n$, the term of the denominator is $a_n + a_1 + a_2$, and $a_{n}$ should be much greater than $a_2, a_1$, so that fraction would be end up being a close number to 1. In a similar way, we let $a_1$ be much greater than $a_2$, $a_3$, so $\frac{a_1}{a_1 + a_2 + a_3}$ is close to 1, and in a similar manner until we reach $n-1$, in which case $a_{n - 1}$ can not be a number much greater than $a_1$, for $a_1$ is the greatest of all numbers, the same explanation applies to case $n$, thus we would have $n-2$ fractions close to $1$ and $2$ fractions of the sum close to cero. With the construction explained above I can easily prove that $lub A_n = n-2, glb A_n = 1$ given that $n-2$ is an upper bound and $1$ a lower bound for $A_n$, but I have not been able to give a proof for such facts. The attempts I have tried is by contradiction, in which follows that, in proving that $n-2$ is an upper bound, there must exist $1 \leq j \leq n$ such that $\frac{a_j}{a_{j}+a_{j+1}+a_{j+2}} > \frac{n-2}{n}$, but nothing I have deduced from that fact. What I would be glad to obtain is an elucidation of what I am not thinking correctly, not a complete proof, for if I wished that, I could check the back of the book for the answer
If anyone is interested, I've managed to solve it. Clue:
To show that $1$ and $n-2$ are lower and upper bounds resp. notice that
$$\frac{a_k}{a_1 + \cdot \cdot \cdot + a_n} < \frac{a_k}{a_k + a_{k+1} + a_{k+2}} = 1-\frac{a_{k+1} + a_{k+2}}{a_k + a_{k+1} + a_{k+2}}$$
and to prove that $glb A_n = 1$, notice that
$$\frac{a_k}{a_k + a_{k+1} + a_{k+2}} < \frac{a_k}{a_k + a_{k+1}} $$
then, for $\epsilon > 0$, there exists $N \in \mathbb{N}$ such that $\epsilon N > 1$, put $a_k = N^{k-1} (n-1)^{k-1}$ for each integer $k$ between $1$ and $n$ closed, consequently use $glb$ definition, similarly for showing $lubA_n = n-2$.