Question 68
Prove that for every positive real numbers $a$, $b$, $c$,
$$
\frac{1+a^2}{1+b} + \frac{1+b^2}{1+c} + \frac{1+c^2}{1+a} \ge 6(\sqrt{2}-1).
$$
$$
\frac{1+a^2}{1+b} + \frac{1+b^2}{1+c} + \frac{1+c^2}{1+a} \ge 6(\sqrt{2}-1).
$$
MathNet 2012 Question 68
Math Olympiad Vault · MathNet 2012