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 $\alpha\leq\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\leq\beta$
 A: 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
$$\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}>1.$$
A: There are no well defined bounds.
Let $S$ denote the given expression. Note first that
$$S > \frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=1.$$
On the other hand, with $a=1$, $c=\varepsilon$ and $b=\varepsilon^2$ we have
$$S=\frac{1}{1+\varepsilon^2} + \frac{2\varepsilon}{1+\varepsilon} \to 1$$
as $\varepsilon \downarrow 0$. Thus the largest lower bound for $S$ is $1$.
For an upper bound, assume wlog that $a \leq b,c$. Then
$$S \leq \frac{a}{a+b} + \frac{b}{b+a}+\frac{c}{c+a} < 2.$$
With $a=1$, $b=\varepsilon$ and $c=\varepsilon^2$ we have
$$S = \frac{2}{1+\varepsilon} + \frac{\varepsilon^2}{1+\varepsilon^2} \to 2$$
as $\varepsilon \downarrow 0$. Hence the smallest upper bound for $S$ is $2$.
