Find $\sup S\;\&\;\inf S$ if exist. $$S:=\left\{\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}:\;a,b,c\in\langle0,+\infty\rangle\right\}$$ My work:
Although it is obvious due to its simmetry, I reformulated the given expression: $$\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}=\frac{1}{1+\frac{b}{a}}+\frac{1}{1+\frac{c}{b}}+\frac{1}{1+\frac{a}{c}}<3\implies\sup S=3$$
$A-G\;\text{mean}$ for possible $\min S$ if $\;\inf S\in S:$ $$\frac{1}{1+\frac{b}{a}}+\frac{1}{1+\frac{c}{b}}+\frac{1}{1+\frac{a}{c}}\geq\frac{3}{\sqrt[3]{\left(1+\frac{b}{a}\right)\left(1+\frac{c}{b}\right)\left(1+\frac{a}{c}\right)}}$$ The equality holds $\iff\;\;a=b=c$ $$\implies \inf S=\frac{3}{2}\in S\implies\min S=\frac{3}{2}$$
Is this correct?
Note: The set was originally named $A$, but I renamed it not to mix it with $A$ for the $\text{arithmetic mean}$.
No, your solution is not correct.
$$\sum_{cyc}\frac{a}{a+b}>\sum_{cyc}\frac{a}{a+b+c}=1.$$ Now, for $c\rightarrow0^+$ we obtain: $$\sum_{cyc}\frac{a}{a+b}\rightarrow\frac{a}{a+b}+1,$$ which for $b\rightarrow+\infty$ is closed to $1$.
Thus, $$\inf\sum_{cyc}\frac{a}{a+b}=1.$$
Also $$\sum_{cyc}\frac{a}{a+b}<\sum_{cyc}\frac{a+c}{a+b+c}=2.$$ Can you end it now?