$abc = 1 \Rightarrow \frac{1}{3}(a+b+c)$. Then:
$$\begin{aligned} c^2 + a + b &\le c^2 + (a+b)\frac{1}{3}(a+b+c) \\ &= \frac{1}{3}(3c^2 + a^2 + b^2 + 2ab + ac + bc) \\ &\le \frac{1}{6}(8c^2 + 5a^2 + 5b^2). \end{aligned}$$
So
$$\frac{a^2 + b^2}{c^2 + a + b} + \dots \ge \frac{a^2 + b^2}{\frac{1}{6}(8c^2 + 5a^2 + 5b^2)} + \dots + \frac{6(a^2 + b^2)}{8c^2 + 5a^2 + 5b^2} + \dots \ge 2,$$
the last inequality must be proved. As three points meant a cyclic permutation of variables. For simplification denote $x = a^2$, $y = b^2$, $z = c^2$. Then we need to prove such inequality:
$$A = \frac{x+y}{8z+5x+5y} + \dots \ge \frac{1}{3}.$$
$$A = \frac{(x+y)^2}{(x+y)(8z+5x+5y)} + \dots \ge \frac{((x+y)^2 + \dots)^2}{(x+y)(8z+5x+5y)} = \frac{4p+8q}{10p+26q} \ge \frac{1}{3},$$
this inequality supervenes from Cauchy-Schwarz inequality. There are used next denote: $p = x^2 + y^2 + z^2$, $q = xy + yz + zx$. The last inequality equivalent $p \ge q$, which are well known.