Bounds on $\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}$ It's simple. What are the bounds on $\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}$ as $a,b,c>0$. Thanks in advance!
Edit: I need bound that can actually be touched i.e. $\alpha$ and $\beta$ such that $\alpha\leq\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\leq\beta$

If $a,\,b,\,c$ are positive real numbers then the answer is $2.$
Indeed, we have
$$\frac{c+a}{a+b+c}-\frac{a}{a+b}=\frac{bc}{(a+b)(a+b+c)} \geqslant 0,$$
so
$$\frac{a}{a+b} \leqslant \frac{c+a}{a+b+c},$$
therefore
$$\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a} \leqslant \frac{(c+a)+(a+b)+(b+c)}{a+b+c}=2.$$
In addition, from
$$\frac{a}{a+b}>\frac{a}{a+b+c},$$
we get
$$\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}>1.$$
