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

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.
