TITLE: on possible generalisations of $1\le\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\le 2$
QUESTION [5 upvotes]: Here is a known double inequality for positive numbers:
$$
1\le\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\le 2
$$
Source: https://www.cut-the-knot.org/m/Algebra/Crux4196.shtml
I'm curious if at least one of them can be generalised in the following way:
$$
C_1(a^{r-1}+b^{r-1}+c^{r-1})\le\frac{a^r}{a+b}+\frac{b^r}{b+c}+\frac{c^r}{c+a}\le C_2(a^{r-1}+b^{r-1}+c^{r-1})
$$
The inequality (or at least one of them, if not possible for both) should be true for any reals $a,b,c>0$, $r\ge 1$;
$C_1,C_2$ are positive numbers not depending on $a,b,c,r$.
I searched a lot, on this site and others like sets of cyclic inequalities e.g. https://www.cut-the-knot.org/m/Algebra/CyclicInequality1.shtml or papers e.g. http://www.m-hikari.com/imf/imf-2018/1-4-2018/p/laiIMF1-4-2018.pdf, but didn't found such result. The only related thing I found is  $  \frac{a}{a+b} + \frac{b}{b+c} + \frac{c}{c+a} \geq  1+\frac{3\sqrt[3]{a \, b \ c}}{2(a+b+c)}$, a stronger version of given inequality, found here:
Cyclic Inequality in n (at least 4) variables

REPLY [2 votes]: The left inequality.
By Chebyshov we obtain:
$$\sum_{cyc}\frac{a^r}{a+b}>\sum_{cyc}\frac{a^r}{a+b+c}\geq\frac{a^{r-1}+b^{r-1}+c^{r-1}}{3}.$$
Id est, we can say that $C_1=\frac{1}{3}.$
Also, for $r=\frac{7}{6}$ by Buffalo Way we can show that the following inequality is true.
$$\sum_{cyc}\frac{a^r}{a+b}\geq\frac{a^{r-1}+b^{r-1}+c^{r-1}}{2}.$$
For $r=\frac{8}{7}$ the last inequality is wrong already.
The right inequality.
$$\sum_{cyc}\frac{a^r}{a+b}<\sum_{cyc}\frac{a^r}{a}=a^{r-1}+b^{r-1}+c^{r-1}$$ and we can say that $C_2=1$.