parmenides51 wrote: Let $a, b, c$ be real numbers such that $ab + bc + ca = 1$. Prove that $$\frac{(a + b)^2 + 1}{c^2+2}+\frac{(b + c)^2 + 1}{a^2+2}+ \frac{(c + a)^2 + 1}{b^2+2} \ge 3$$ 2010 Romania JBMO here

$\sum{\frac{(a+b)^2}{c^2+2}}+\sum{\frac{1}{c^2+2}} \geq 3$ By $Titu$ $\frac{4(a+b+c)^2+9}{a^2+b^2+c^2+6} \geq 3$ $a^2+b^2+c^2 \geq 1$ And we are done