<s>[INST] Prove $\frac{1+a^{2}}{1+b+c^{2}} +\frac{1+b^{2}}{1+c+a^{2}} +\frac{1+c^{2}}{1+a+b^{2}} \geq 2$ For positive integers Numbers  $a, b, c $ prove that  $$\frac{1+a^{2}}{1+b+c^{2}} +\frac{1+b^{2}}{1+c+a^{2}} +\frac{1+c^{2}}{1+a+b^{2}} \geq  2$$
This inequality above take long time to prove it and still couldn't complete it.  How I can prove this inequality?  Any hint will help. Thanks
 [/INST] Since 
$$\sum_{cyc}(1+a^2)(1+b+c^2)\sum_{cyc}\frac{1+a^2}{1+b+c^2} \ge \left(\sum_{cyc} (1+a^2)\right)^2,$$
we only need to show that 
$$\left(\sum_{cyc} (1+a^2)\right)^2\ge 2\sum_{cyc}(1+a^2)(1+b+c^2).\tag{1}$$
The LHS is
$$\begin{aligned}2\sum_{cyc}(1+a^2+b+a^2b+c^2+a^2c^2)&= 6+4(a^2+b^2+c^2)+2(a^2b+b^2c+c^2a)\\
&+2(a+b+c)+2(a^2c^2+c^2b^2+b^2a^2).
\end{aligned}\tag{2}
$$
The RHS is
$$(3+a^2+b^2+c^2)^2=9+6(a^2+b^2+c^2)+(a^2+b^2+c^2)^2.\tag{3}$$
Compare (2) and (3), we see that (1) is equivalent to
$$a^4+b^4++c^4+2(a^2+b^2+c^2)+3\ge 2(a^2b+b^2c+c^2a)+2(a+b+c).\tag{4}$$
The last inequality is obvious from AM-GM:
$$a^4+b^2\ge 2a^2b,\dots$$
and
$$a^2+1\ge 2a,\dots$$
To finish the proof, we note that the equality happens in (4) only when $a=b=c=1$, which also yields equality 
$$\sum_{cyc}\frac{1+a^2}{1+b+c^2}=2.$$
</s>