$\frac{a}{1+b^2}=a-\frac{ab^2}{1+b^2}\geq a-\frac{ab^2}{2b}=a-\frac{ab}{2}$Tương tự $\frac{b}{1+c^2}\geq b- \frac{bc}{2};\frac{c}{1+a^2}\geq c - \frac{ca}{2}$
$\Leftrightarrow \frac{a}{1+b^2}+\frac{b}{1+c^2}+\frac{c}{1+a^2} \geq a+b+c - \frac{ab+bc+ca}{2} \geq 3- \frac{3}{2}=\frac{3}{2}$