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?
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 $x\to 0^+$).