<s>[INST] For $abc=1$ prove that $\sum\limits_\text{cyc}\frac{1}{a+3}\geq\sum\limits_\text{cyc}\frac{a}{a^2+3}$ Let $a$, $b$ and $c$ be positive numbers such that $abc=1$. Prove that:
$$\frac{1}{a+3}+\frac{1}{b+3}+\frac{1}{c+3}\geq\frac{a}{a^2+3}+\frac{b}{b^2+3}+\frac{c}{c^2+3}$$
I tried TL, BW, the Vasc's Theorems and more, but without success.
I proved this inequality! 
I proved also the hardest version: $\sum\limits_{cyc}\frac{1}{a+4}\geq\sum\limits_{cyc}\frac{a}{a^2+4}$. 
Thanks all!
 [/INST] Another way.
Since for any $a>0$ we have $$\frac{1}{a+3}-\frac{a}{a^2+3}+\frac{9}{64}\geq\frac{27}{64\left(a^{\frac{8}{3}}+a^{\frac{4}{3}}+1\right)}$$ and for positives $a$, $b$ and $c$ such that $abc=1$ we have $$\sum_{cyc}\frac{1}{a^2+a+1}\geq1,$$ we obtain:
$$\sum_{cyc}\left(\frac{1}{a+3}-\frac{a}{a^2+3}\right)=\sum_{cyc}\left(\frac{1}{a+3}-\frac{a}{a^2+3}+\frac{9}{64}\right)-\frac{27}{64}\geq$$
$$\geq\frac{27}{64}\left(\sum_{cyc}\frac{1}{a^{\frac{8}{3}}+a^{\frac{4}{3}}+1}-1\right)\geq0.$$
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