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
 A: The above inequality holds for any positive real numbers $a,b,c$.
Indeed, denote $x=1+b+c^2,y=1+c+a^2,z=1+a+b^2$, the inequality becomes
$$\frac{y-c}{x} + \frac{z-a}{y} + \frac{x-b}{z} \ge 2,$$
or equivalently 
$$\frac{y}{x} + \frac{z}{y} + \frac{x}{z} \ge 2 + \frac{c}{x} + \frac{a}{y} + \frac{b}{z}.$$
By AM-GM inequality:
$$\frac{y}{x} + \frac{z}{y} + \frac{x}{z} \ge \sqrt[3]{\frac{y}{x} \frac{z}{y} \frac{x}{z}} =3.$$
It remains to prove $$\frac{c}{x} + \frac{a}{y} + \frac{b}{z} \le 1.$$
Since $x\ge 2c+b,y\ge 2a+c,z\ge 2b+a$ we only need to prove that
$$\frac{c}{2c+b} + \frac{a}{2a+c} + \frac{b}{2b+a} \le 1,$$ which is equivalent to
$$\left(\frac{c}{2c+b} - \frac{1}{2}\right) + \left(\frac{a}{2a+c} - \frac{1}{2}\right) + \left(\frac{b}{2b+a} - \frac{1}{2}\right) \le 1 - \frac{3}{2}$$
or 
$$\frac{b}{2c+b} + \frac{c}{2a+c} + \frac{a}{2b+a} \ge 1.$$
The last inequality is true by Cauchy-Schwarz inequality:
$$\sum\frac{a}{2b+a} = \sum\frac{a^2}{2ab+a^2} \ge \frac{(a+b+c)^2}{\sum (2ab+a^2)}=1.$$
We are done.
A: 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 
$$\sum_{cyc}\frac{1+a^2}{1+b+c^2}=2.$$
A: Let $1+a^2=x$, $1+b^2=y$ and $1+c^2=z$. 
Hence, by AM-GM and C-S we obtain:
$$\sum_{cyc}\frac{a^2+1}{1+b+c^2}\geq\sum_{cyc}\frac{a^2+1}{1+\frac{1+b^2}{2}+c^2}=\sum_{cyc}\frac{2x}{y+2z}=\sum_{cyc}\frac{2x^2}{xy+2xz}\geq$$
$$\geq\frac{2(x+y+z)^2}{\sum\limits_{cyc}(xy+2xz)}=\frac{2(x+y+z)^2}{3(xy+xz+yz)}=2+\frac{\sum\limits_{cyc}(x-y)^2}{3(xy+xz+yz)}\geq2.$$
Done!
