<s>[INST] $\inf$ and $\sup$ of a set given by $\sum\limits_{k=1}^{n}\frac{a_{k}}{a_{k}+a_{k+1}+a_{k+2}}$ Let $n\geq3$ be an arbitrarily fixed integer. Take all the possible finite sequences $(a_{1},...,a_{n})$ of positive numbers. Find the supremum and the infimum of the set of numbers  $$\sum_{k=1}^{n}\frac{a_{k}}{a_{k}+a_{k+1}+a_{k+2}},$$ where we put $a_{n+1}=a_{1}$ and $a_{n+2}=a_{2}$.
Attempted solution:
$$n=3\rightarrow       \sum_{k=1}^{n}\frac{a_{k}}{a_{k}+a_{k+1}+a_{k+2}}=1$$
$$n>3\wedge a_{i}>0\rightarrow \frac{a_{k}}{\sum_{i=1}^{n}a_{i}}<  \frac{a_{k}}{a_{k}+a_{k+1}+a_{k+2}}<  \frac{\sum_{i=1}^{k}a_{i}+\sum_{i=k+3}^{n}a_{i}}{\sum_{i=1}^{n}a_{i}}$$
$$\rightarrow \sum_{k=1}^{n} \frac{a_{k}}{\sum_{i=1}^{n}a_{i}}<  \sum_{k=1}^{n}\frac{a_{k}}{a_{k}+a_{k+1}+a_{k+2}}<\sum_{k=1}^{n}  \frac{\sum_{i=1}^{n}a_{i}-a_{k+1}-a_{k+2}}{\sum_{i=1}^{n}a_{i}}$$
$$\rightarrow 1< \sum_{k=1}^{n}\frac{a_{k}}{a_{k}+a_{k+1}+a_{k+2}}< n-2$$
I don't see why these must be the tightest bounds. Can $\inf$ and $\sup$ be calculated from these inequalities?
 [/INST] Your bounds are sharp (to my surprise).
Example: $a_k = x^k$.  Then
\begin{align*}
\sum_{k=1}^n \frac{a_k}{a_k + a_{k+1} + a_{k+2}}
&= \sum_{k=1}^{n-2} \frac{x^k}{x^k + x^{k+1} + x^{k+2}}
   + \frac{x^{n-1}}{x^{n-1} + x^n + x}
   + \frac{x^n}{x^n + x + x^2} \\
&= \frac{n-2}{1 + x + x^2} + \frac{1}{1 + x + x^{-n+1}} + \frac{1}{1 + x^{-n+1} + x^{-n+2}}
\end{align*}
This tends to $1$ as $x\to\infty$ and to $n-2$ as $x\to 0^+$.
The only essential feature of the functions $x^k$ here is that each grows faster than the previous one (as $x\to\infty$) and shrinks faster than the previous one (as $x\to 0^+$).
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