$a,b,c,d>0$ and $abcd=1$ prove $\sum \frac {1}{(1+a)(1+a^2)} \ge 1$

Question

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.

Answer 1

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.