, 2012

Solution 1. Using AM-GM inequality we have
$$\frac{1+a^2}{1+b} + \frac{1+b^2}{1+c} + \frac{1+c^2}{1+a} \ge 3\sqrt[3]{\frac{1+a^2}{1+a} \cdot \frac{1+b^2}{1+b} \cdot \frac{1+c^2}{1+c}} \quad (1)$$
On the other hand, for every positive real number $x$, the following inequality holds:
$$\frac{1+x^2}{1+x} \ge 2(\sqrt{2}-1). \quad (2)$$
Indeed, the inequality (2) is equivalent to
$$x^2 - 2(\sqrt{2}-1)x + 1 - (2\sqrt{2}-2) \ge 0,$$
i.e.
$$x^2 - 2(\sqrt{2}-1)x + (\sqrt{2}-1)^2 \ge 0,$$
and hence $(x - (\sqrt{2} - 1))^2 \ge 0$.
From (1) and (2) we obtain
$$\frac{1+a^2}{1+b} + \frac{1+b^2}{1+c} + \frac{1+c^2}{1+a} \ge 3\sqrt[3]{8(\sqrt{2}-1)^3} = 6(\sqrt{2}-1).$$
We have equality if and only if $a = b = c = \sqrt{2} - 1$.

Solution 2. The triples
$$(1+a^2, 1+b^2, 1+c^2) \quad \text{and} \quad \left( \frac{1}{1+a}, \frac{1}{1+b}, \frac{1}{1+c} \right)$$
have opposite monotony. From the Rearrangement Inequality, it follows
$$\sum_{cyc} \frac{1+a^2}{1+b} \ge \sum \frac{1+a^2}{1+a} \ge 6(\sqrt{2}-1),$$
where we have used the inequality (2) in the previous solution.