<s>[INST] $a,b,c,d>0$ and $abcd=1$ prove $\sum \frac {1}{(1+a)(1+a^2)} \ge 1$ If $a,b,c,d>0$ and $abcd=1$ prove: 
$$\sum \frac{1}{(1+a)(1+a^2)}\ge 1$$
Here's my solution:
I try to prove it by reverse:
$$3\ge\sum\frac{a^3+a^2+a}{(a+1)(a^2+1)}$$Then by AM-GM:
$$3\ge\sum\frac{a(1+a+a^2)}{2a(a+1)}$$$$$6\ge\sum\frac{1+a+a^2}{a+1}$$
Then We need to prove:
$$2\ge\sum\frac{a^2}{a+1}$$
This inequality but I still can't prove it. 
I don't know the last inequality is right or wrong.
There's another solution from another user @Michael Rozenberg by Vasc's RCF theorem.
 [/INST] The inequality $\sum\limits_{cyc}\frac{a^2}{a+1}\leq2$ is wrong.
Try $b=c=d\rightarrow0^+$ and $a\rightarrow+\infty$.
The starting inequality we can prove also by the following two inequalities.
$$\frac{1}{(a^2+1)(a+1)}+\frac{1}{(b^2+1)(b+1)}\geq\frac{1}{1+\sqrt{a^3b^3}}$$ and
$$\frac{1}{(c^2+1)(c+1)}+\frac{1}{(d^2+1)(d+1)}\geq\frac{1}{1+\sqrt{c^3d^3}}$$
and the rest is smooth. 
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