Prove $\frac{a}{b^{2}+1} + \frac{b}{c^{2}+1} + \frac{c}{a^{2} + 1} \ge \frac{3}{2}$ $a,b,c > 0$ and $a+b+c=3$, prove
$$ \frac{a}{b^{2} + 1} + \frac{b}{c^{2}+1} + \frac{c}{a^{2}+1} \ge 3/2 $$

Attempt:
Notice that by AM-Gm
$$\frac{a}{b^{2} + 1} + \frac{b}{c^{2}+1} + \frac{c}{a^{2}+1} \ge  3\frac{\sqrt[3]{abc}}{\sqrt[3]{(a^{2}+1)(b^{2}+1)(c^{2}+1)}} $$
Now, AM-GM again 
$$ a^{2}+b^{2}+c^{2} + 3 \ge 3 \sqrt[3]{(a^{2}+1)(b^{2}+1)(c^{2}+1)} ... (1)$$
Then $a+b+c = 3 \ge 3 \sqrt[3]{abc} \implies 1 \ge \sqrt[3]{abc}$. Also
$$a^{2}  + b^{2} + c^{2} \ge 3 \sqrt[3]{(abc)^{2}}$$
multiply by $1 \ge \sqrt[3]{abc}$ and will get
$$ a^{2}  + b^{2} + c^{2} \ge 3 abc ... (2)$$
subtract $(1)$ with $(2)$ and get
$$ 3 \ge 3 \sqrt[3]{(a^{2}+1)(b^{2}+1)(c^{2}+1)} - 3 abc$$
$$ 3 + 3 abc \ge \sqrt[3]{(a^{2}+1)(b^{2}+1)(c^{2}+1)} $$
$$ \frac{3abc}{\sqrt[3]{(a^{2}+1)(b^{2}+1)(c^{2}+1)}} \ge 1 - \frac{3}{\sqrt[3]{(a^{2}+1)(b^{2}+1)(c^{2}+1)}} $$
How to continue..?
 A: $$\sum_{cyc}\frac{a}{b^2+1}=3+\sum_{cyc}\left(\frac{a}{b^2+1}-a\right)=3-\sum_{cyc}\frac{ab^2}{b^2+1}\geq$$
$$\geq3-\sum_{cyc}\frac{ab^2}{2b}=3-\frac{1}{2}(ab+ac+bc).$$
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:
$$a^2+b^2+c^2=\frac{1}{3}(1^2+1^2+1^2)(a^2+b^2+c^2)\geq\frac{1}{3}(a+b+c)^2=3.$$
A: 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 verify. :)
A: Another way.
Your inequality is a sixth degree. We can reduce this degree by the Bacteria's method.
Indeed, by C-S, Murhead, Rearrangement and SOS(here it's also a Tangent Line method) we obtain: $$\sum_{cyc}\frac{a}{b^2+1}=\sum_{cyc}\frac{a^2(a+c)^2}{a(a+c)^2(b^2+1)}\geq\frac{\left(\sum\limits_{cyc}(a^2+ab)\right)^2}{\sum\limits_{cyc}a(a+c)^2(b^2+1)}=$$
$$=\tfrac{3(a+b+c)\left(\sum\limits_{cyc}(a^2+ab)\right)^2}{\sum\limits_{cyc}a(a+c)^2(9b^2+(a+b+c)^2)}=\frac{3}{2}+\tfrac{3\left(2(a+b+c)\left(\sum\limits_{cyc}(a^2+ab)\right)^2-\sum\limits_{cyc}a(a+c)^2(9b^2+(a+b+c)^2)\right)}{2\sum\limits_{cyc}a(a+c)^2(9b^2+(a+b+c)^2)}=$$
$$=\frac{3}{2}+\frac{3\sum\limits_{cyc}(a^5+3a^4b+2a^4c-4a^3b^2+4a^3c^2+8a^3bc-14a^2b^2c)}{2\sum\limits_{cyc}a(a+c)^2(9b^2+(a+b+c)^2)}=$$
$$=\frac{3}{2}+\tfrac{3\sum\limits_{cyc}(a^5+3a^4b+2a^4c-4a^3b^2-2a^3c^2+8(a^3bc-a^2b^2c)+6(a^3c^2-a^2b^2c))}{2\sum\limits_{cyc}a(a+c)^2(9b^2+(a+b+c)^2)}\geq$$
$$\geq\frac{3}{2}+\frac{3\sum\limits_{cyc}(a^5+3a^4b+2a^4c-4a^3b^2-2a^3c^2)}{2\sum\limits_{cyc}a(a+c)^2(9b^2+(a+b+c)^2)}=$$
$$=\frac{3}{2}+\frac{3\sum\limits_{cyc}(a^5+3a^4b-4a^3b^2-2a^2b^3+2ab^4)}{2\sum\limits_{cyc}a(a+c)^2(9b^2+(a+b+c)^2)}=$$
$$=\frac{3}{2}+\frac{3\sum\limits_{cyc}a(a-b)(a^3+4a^2b-2b^3)}{2\sum\limits_{cyc}a(a+c)^2(9b^2+(a+b+c)^2)}=$$
$$=\frac{3}{2}+\frac{3\sum\limits_{cyc}\left(a(a-b)(a^3+4a^2b-2b^3)-\frac{3(a^5-b^5)}{5}\right)}{2\sum\limits_{cyc}a(a+c)^2(9b^2+(a+b+c)^2)}=$$
$$=\frac{3}{2}+\frac{3\sum\limits_{cyc}(a-b)^2(2a^3+19a^2b+16ab^2+3b^3)}{10\sum\limits_{cyc}a(a+c)^2(9b^2+(a+b+c)^2)}\geq\frac{3}{2}.$$
