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!
 A: BW in the following version does not help.
Let $a=x^3$, $b=y^3$ and $c=z^3$.
Hence, we need to prove that
$$\sum_{cyc}\frac{1}{x^3+3xyz}\geq\sum_{cyc}\frac{x^3}{x^6+3x^2y^2z^2}$$ or
$$\sum_{cyc}\frac{1}{x^3+3xyz}\geq\sum_{cyc}\frac{x}{x^4+3y^2z^2}.$$
Now, we can assume that $x=\min\{x,y,z\}$, $y=x+u$ and $z=x+v$ 
and these substitutions give inequality, which I don't know to prove.
But we can use another BW!
Let $a=\frac{y}{x}$, $b=\frac{z}{y}$ and $c=\frac{x}{z}$, where $x$, $y$ and $z$ are positives. 
Hence, we need to prove that
$$\sum_{cyc}\frac{x}{3x+y}\geq\sum_{cyc}\frac{xy}{3x^2+y^2}$$ or
$$\sum_{cyc}\frac{x^3-x^2y}{(3x+y)(3x^2+y^2)}\geq0.$$
Now, let $x=\min\{x,y,z\}$, $y=x+u$ and $z=x+v$.
Hence, we need to prove that
$$128(u^2-uv+v^2)x^7+16(16u^3+23u^2v-15uv^2+16v^3)x^6+$$
$$+32(8u^4+27u^3v+12u^2v^2-11uv^3+8v^4)x^5+$$
$$+4(32u^5+193u^4v+266u^3v^2-42u^2v^3-33uv^4+32v^5)x^4+$$
$$+2(8u^6+178u^5v+435u^4v^2+152u^3v^3-99u^2v^4+30uv^5+8v^6)x^3+$$
$$+uv(45u^5+375u^4v+291u^3v^2-83u^2v^3+57uv^4+3v^5)x^2+$$
$$+2u^2v^2(24u^4+66u^3v-18u^2v^2+13uv^3+3v^4)x+$$
$$+u^3v^3(18u^3-6u^2v+3uv^2+v^3)\geq0,$$
which is obvious.
Done!
A: 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:
$$\sum_{cyc}\left(\frac{1}{a+3}-\frac{a}{a^2+3}\right)=\sum_{cyc}\left(\frac{1}{a+3}-\frac{a}{a^2+3}+\frac{9}{64}\right)-\frac{27}{64}\geq$$
$$\geq\frac{27}{64}\left(\sum_{cyc}\frac{1}{a^{\frac{8}{3}}+a^{\frac{4}{3}}+1}-1\right)\geq0.$$
A: Partial Hint and too long for a comment :
Case where two varaibles are superior two one .
If we define :
$$f\left(x\right)=\frac{1}{e^{x}+3}-\frac{e^{x}}{e^{2x}+3},g\left(x\right)=\ln\left(1+e^{-x}\right)-\ln\left(2\right)$$
Show that for $x\in(-\infty,\infty)$ :
$$f''(x)+g''(x)>0$$
Now apply Jensen's inequality on two variable .
Remains to show :
$$-\left(g\left(a\right)+g\left(b\right)-2g\left(\frac{-c}{2}\right)\right)+f\left(c\right)\geq 0$$
Or :
$$-\left(g\left(a\right)+g\left(b\right)-2g\left(\frac{-c}{2}\right)\right)+\ln\left(f\left(c\right)+1\right)\geq 0$$
Wich is true for $a+b+c=0, a,b\in(0,2) \operatorname{or} a,b\in(2,\infty)$,


Last hint remark  with the constraint  $x,y\in [1,\infty)$:
$$\frac{\left(1+\frac{1}{xy+3}-\frac{xy}{\left(xy\right)^{2}+3}\right)\left(1+\sqrt{xy}\right)^{2}}{\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)}\ge \frac{\left(3+\frac{1}{xy+3}-\frac{xy}{\left(xy\right)^{2}+3}\right)\left(3+\sqrt{xy}\right)^{2}}{3\left(3+\frac{1}{x}\right)\left(3+\frac{1}{y}\right)}\ge \frac{4\left(3+\frac{1}{xy+3}-\frac{xy}{\left(xy\right)^{2}+3}\right)\left(3+\sqrt{xy}\right)}{3\left(3+\frac{1}{x}\right)\left(3+\frac{1}{y}\right)}\ge1$$
Case where two variable are less than one
Now consider the function :
$$p\left(x\right)=\frac{1}{x+3}-\frac{x}{x^{2}+3}$$
There exists a positive constant $C$ and $b\in(0,1),x>0$ such that :
$$p\left(Cx\right)-p\left(Cx^{b}\right)\geq 0$$
$$C=\left(3-2\sqrt{2}\right)^{\frac{1}{3}}+\left(3+2\sqrt{2}\right)^{\frac{1}{3}}$$
To be continued...
