Solution verification: $\sup S\;\&\;\inf S,\;S:=\left\{\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}:\;a,b,c\in\langle 0,+\infty\rangle\right\}$ 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?
