<s>[INST] If $abc=1$, prove that $\frac{2}{(a+1)^2+b^2+1} + \frac{2}{(b+1)^2+c^2+1} + \frac{2}{(c+1)^2+a^2+1} \le 1$. If $a, b, c \in \mathbb{N}$, and $abc = 1$, prove that:
$$S = \frac{2}{(a+1)^2+b^2+1} + \frac{2}{(b+1)^2+c^2+1} + \frac{2}{(c+1)^2+a^2+1} \le 1$$
Here is my try:
$$\begin{align}\frac{2}{(a+1)^2+(b^2+1)} & \le \frac{1}{\sqrt{(a+1)^2 \cdot (b^2+1)}}\\
& = \frac{1}{(a+1) \cdot \sqrt{b^2+1}}\\
& \le \frac{1}{(a+1) \cdot b}\\
& \le \frac{1}{ab}
\end{align}$$
So,
$$S \le \frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca} = \frac{a+b+c}{1}$$
Can you help me to continue it, please? Thanks!
 [/INST] This is a problem from Olympiad Inequalities, Thomas J. Mildorf, 2005 (short but wonderful reading with many different ideas, a true gem). The idea presented here (replacing $a,b,c$ with $x/y,y/z,z/x$) is often used to normalize the inequality in case of $abc=1$. 

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