Question 126
Prove that for non-negative real numbers $a$ and $b$ such that $a + b \le 2$ the inequality
$$
\frac{1}{1+a^2} + \frac{1}{1+b^2} \le \frac{2}{1+ab}
$$
holds. When is the equality attained? (Austria 2018)
$$
\frac{1}{1+a^2} + \frac{1}{1+b^2} \le \frac{2}{1+ab}
$$
holds. When is the equality attained? (Austria 2018)