<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] 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!
</s>