Supremum and infimum of a set of numbers $\sum_{k=1}^n{\frac{a_k}{a_k+a_{k+1}+a_{k+2}}}$ Let $n \geq 3$ be an arbitrarily fixed integer. Take all the possible finite sequences $(a_1,...,a_n)$ of positive numbers. Find the supremum and infimum of the set of numbers $$\sum_{k=1}^n{\frac{a_k}{a_k+a_{k+1}+a_{k+2}}}$$ where $a_{n+1}=a_1$ and $a_{n+2}=a_2$.
I try to formulate an inequality for the sum, that is $?\leq \sum_{k=1}^n{\frac{a_k}{a_k+a_{k+1}+a_{k+2}}} \leq ?$. But I have no idea. Can anyone give some hints?
 A: (Write $S(a)=\sum \dots$ for any such sum)
Plug in $q,q^2,\dots,q^n$ for a $q > 0$ and get
$$
S(a) = (n-2)\frac{1}{1+q+q^2}+\frac{q^{n-1}}{q^{n-1}+q^n+q^1}+\frac{q^{n}}{q^{n}+q^1+q^2}
$$
let $q \rightarrow 0,\infty$ yield inf $\leq 1$ and $\sup\geq n-2$
Not let $a_1,\dots,a_n$ be any such sequence, let $A:=\sum a_i$
1) make the denominators bigger:
$$
S(a) \geqslant \frac{a_1}{A}+\dots+\frac{a_n}{A} = \frac{A}{A} = 1
$$
2) observe that for any $x>0$ and quotient $\frac{a}{b}$ where $0 < a < b$
$$
\frac{a}{b} \leqslant \frac{a+x}{b+x} 
$$
because $a(b+x)\leqslant b(a+x)$ is equivalent to $ax\leqslant bx$
3) apply 2) picking $x$ such that the sum in the denominator becomes $A$.
$$
S(a)=\sum \frac{a_i}{a_i+a_{i+1}+a_{i+2}}
\leq
\sum \frac{a_i+\left[A-(a_i+a_{i+1}+a_{i+2})\right]}{a_i+a_{i+1}+a_{i+2}+\left[A-(a_i+a_{i+1}+a_{i+2})\right]}$$
and this is
$$
=\sum \frac{A-(a_{i+1}+a_{i+2})}{A} = 
\frac{nA-a_2-a_3-a_3-a_4-a_4-a_5-\dots-a_{n-1}-a_n-a_n-a_1-a_1-a_2}{A}
$$
Here every $a_i$ appears exactly twice and therefore $S(a) \leqslant n-2$
