Saudi Arabia Mathematical Competitions 2012

Solution 1. Using AM-GM inequality we have $$\frac{1+a^2}{1+b}+\frac{1+b^2}{1+c}+\frac{1+c^2}{1+a}\ge 3\sqrt[3]{\frac{1+a^2}{1+a}\cdot \frac{1+b^2}{1+b}\cdot \frac{1+c^2}{1+c}}(1)$$ On the other hand, for every positive real number $x$, the following inequality holds: $$\frac{1+x^2}{1+x}\ge 2(\sqrt{2}-1).(2)$$ 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 $$(1+a^2,1+b^2,1+c^2)\text{and}\left(\frac{1}{1+a},\frac{1}{1+b},\frac{1}{1+c}\right)$$ 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.