Show that, for all positive real numbers $a, b, c$ such that $abc = 1$, the inequality $$\frac{1}{1 + a^2 + (b + 1)^2} +\frac{1}{1 + b^2 + (c + 1)^2} +\frac{1}{1 + c^2 + (a + 1)^2} \le \frac{1}{2}$$
Source: 2012 Romania JBMO TST 4.1
Tags: inequalities, algebra
Show that, for all positive real numbers $a, b, c$ such that $abc = 1$, the inequality $$\frac{1}{1 + a^2 + (b + 1)^2} +\frac{1}{1 + b^2 + (c + 1)^2} +\frac{1}{1 + c^2 + (a + 1)^2} \le \frac{1}{2}$$