<s>[INST] Calculate the maximum value of $\sum_{cyc}\frac{bc}{(b + c)^3(a^2 + 1)} $ where $a, b, c \in \mathbb R^+$ satisfying $abc = 1$. 
Calculate the maximum value of $$\large \frac{bc}{(b + c)^3(a^2 + 1)} + \frac{ca}{(c + a)^3(b^2 + 1)} + \frac{ab}{(a + b)^3(c^2 + 1)}$$ where $a, b, c$ are positives satisfying $abc = 1$.

We have that $$\sum_{cyc}\frac{bc}{(b + c)^3(a^2 + 1)} \le \frac{1}{2} \cdot \sum_{cyc}\frac{1}{(b + c)(a^2 + 1)} \le \sum_{cyc}\frac{1}{(b + c)(a + 1)^2}$$
$$ = \sum_{cyc}\frac{1}{a(ab + ca + b + c)(bc + 1)} \le \dfrac{1}{2} \cdot \sum_{cyc}\frac{1}{a\sqrt{bc}(ab + ca + 2\sqrt{bc})}$$
$$ = \frac{1}{2} \cdot \sum_{cyc}\frac{1}{a\sqrt a(b + c) + 2}$$
That's all I got, not because I can't go for more, but since I went overboard with this, there's no use trying to push for more.
 [/INST] The inequality $$\sum_{cyc}\frac{x}{x^4+1}\leq\frac{3}{2}$$ for positives $x$, $y$ and $z$ we can prove also by the following way.
We'll prove that
$$\frac{x}{x^4+1}\leq\frac{3(x^2+1)}{4(x^4+x^2+1)}$$ is true for all positive $x$.
Indeed, let $x^2+1=2ux.$
Thus, by AM-GM $u\geq1$ and it's enough to prove that
$$3\cdot2u\cdot(4u^2-2)\geq4(4u^2-1)$$ or
$$6u^3-4u^2-3u+1\geq0$$ or $$(u-1)(6u^2+2u-1)\geq0,$$ which is obvious.
Id est, it's enough to prove that
$$\sum_{cyc}\frac{x^2+1}{x^4+x^2+1}\leq2$$ or
$$\sum_{cyc}\frac{x+1}{x^2+x+1}\leq2$$ or
$$\sum_{cyc}\left(\frac{x+1}{x^2+x+1}-1\right)\leq-1$$ or
$$\sum_{cyc}\frac{x^2}{x^2+x+1}\geq1$$ or
$$\sum_{cyc}(x^2y^2-x)\geq0$$ or
$$\sum_{cyc}(x^2y^2-x^2yz)\geq0$$ or
$$\sum_{cyc}z^2(x-y)^2\geq0$$ and we are done!
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