SOLUTION. The inequality is equivalent to (homogeneous form)
$$\frac{a+b}{b+\frac{a+b+c}{3}}+\frac{b+c}{c+\frac{a+b+c}{3}}+\frac{c+a}{a+\frac{a+b+c}{3}} \geq 3$$

Since this problem is cyclic, homogeneous, and holds if $a=b=c$, we can assume that $c=0$ and $a+b=3$. In this case, we have to prove that
$$\begin{array}{l} \frac{3}{1+b}+b+\frac{a}{1+a} \geq 3 \\ \Leftrightarrow \frac{3}{4-a}+\frac{a}{1+a} \geq a \\ \Leftrightarrow a^{3}-4 a^{2}+3 a+3 \geq 0 \\ \Leftrightarrow a(a-2)^{2}+(3-a) \geq 0 \end{array}$$
which is obvious because $a \in[0,3]$. The equality holds for only one case, $a=b=c=1$.