Prove $\frac{a}{b^{2}+1} + \frac{b}{c^{2}+1} + \frac{c}{a^{2} + 1} \ge \frac{3}{2}$

$$\sum_{cyc}\frac{a}{b^2+1}=3+\sum_{cyc}\left(\frac{a}{b^2+1}-a\right)=3-\sum_{cyc}\frac{ab^2}{b^2+1}\geq$$ $$\geq3-\sum_{cyc}\frac{ab^2}{2b}=3-\frac{1}{2}(ab+ac+bc).$$ Can you end it now?

Since by your work $$3-\frac{1}{2}(ab+ac+bc)=3-\frac{1}{2}\cdot\frac{9-a^2-b^2-c^2}{2},$$ it's enough to prove that $$3-\frac{1}{2}\cdot\frac{9-a^2-b^2-c^2}{2}\geq\frac{3}{2}$$ or $$a^2+b^2+c^2\geq3,$$ which is true by C-S: $$a^2+b^2+c^2=\frac{1}{3}(1^2+1^2+1^2)(a^2+b^2+c^2)\geq\frac{1}{3}(a+b+c)^2=3.$$

Since for $x>0$ we have (just draw a graph for ${1\over 1+x^2}$ and a tangent at $x=1$) $${1\over 1+x^2}\geq -{1\over 2}x+1$$ it is enough to check if $$-{1\over 2}(ab+bc+ca)+3\geq {3\over 2}$$ i.e. $$3\geq ab+bc+ca$$ is true?

Since $$a^2+b^2+c^2\geq ab+bc+ca$$ that is easy to verfy. :)