<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] Let $a=b=c=1$.
Thus, we obtain a value $\frac{3}{16}.$
We'll prove that it's a maximal value.
Indeed, by AM-GM  $$\sum_{cyc}\frac{bc}{(b+c)^3(a^2+1)}\leq\sum_{cyc}\frac{bc}{8\sqrt{b^3c^3}(a^2+1)}=\sum_{cyc}\frac{\sqrt{a}}{8(a^2+1)}.$$
Id est, it's enough to prove that
$$\sum_{cyc}\frac{a}{a^4+1}\leq\frac{3}{2},$$ where $a$, $b$ and $c$ are positives such that $abc=1$. 
We have $$\frac{3}{2}-\sum_{cyc}\frac{a}{a^4+1}=\sum_{cyc}\left(\frac{1}{2}-\frac{a}{a^4+1}-\frac{1}{2}\ln{a}\right).$$
Let $f(x)=\frac{1}{2}-\frac{x}{x^4+1}-\frac{1}{2}\ln{x}.$
Thus, easy to see that $f(x)\geq0$ for all $0<x<2$ 
and from here our inequality is proven for $\max\{a,b,c\}<2$.
Let $a\geq2$.
Thus, by AM-GM:
$$\sum_{cyc}\frac{a}{a^4+1}\leq\frac{2}{2^4+1}+2\cdot\frac{x}{4\sqrt[4]{x^4\cdot\left(\frac{1}{3}\right)^3}}=\frac{2}{17}+\frac{\sqrt[4]{27}}{2}<\frac{3}{2}$$ and we are done!
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