Prove that $\sum\limits_{cyc}\frac{a}{a^2+ab+b^2+3}\leq\frac{1}{2}$ 
Let $a$, $b$ and $c$ be non-negative numbers. Prove that:
  $$\frac{a}{a^2+ab+b^2+3}+\frac{b}{b^2+bc+c^2+3}+\frac{c}{c^2+ca+a^2+3}\leq\frac{1}{2}$$

I think this inequality is very interesting because most of the contest's inequalities are homogeneous, 
but this inequality is non-homogeneous.
Testing for $c=0$ gives
$$\frac{a}{a^2+ab+b^2+3}+\frac{b}{b^2+3}\leq0.455...<\frac{1}{2}.$$
For $b=c=0$ we obtain something obvious.
One of the standard ways to prove these inequalities is to try to make a homogenization. 
By the way, trying of homogenization gives a wrong inequality:
$$\sum\limits_{cyc}\frac{a}{a^2+ab+b^2+3}\leq\sum\limits_{cyc}\frac{a}{2\sqrt{3(a^2+ab+b^2}}.$$
Thus, it remains to prove that
$$\sum\limits_{cyc}\frac{a}{\sqrt{3(a^2+ab+b^2)}}\leq1,$$
which is wrong for $c=0$ and $a\rightarrow+\infty$.
Also we can try the following.
We know that $\sum\limits_{cyc}\frac{x}{2x+y}\le1$ for positives $x$, $y$ and $z$.
Indeed, by C-S we obtain:
$$1-\sum_{cyc}\frac{x}{2x+y}=1-\frac{3}{2}-\sum_{cyc}\left(\frac{x}{2x+y}-\frac{1}{2}\right)=\frac{1}{2}\sum_{cyc}\frac{y}{2x+y}-\frac{1}{2}=$$
$$=\frac{1}{2}\sum_{cyc}\frac{y^2}{2xy+y^2}-\frac{1}{2}\geq\frac{1}{2}\frac{(x+y+z)^2}{\sum\limits_{cyc}(y^2+2xy)}=\frac{1}{2}-\frac{1}{2}=0.$$
Thus, it's enough to prove that
$$a^2+ab+b^2+3\geq2(2a+\sqrt{ab})$$ 
because if it's true so 
$$\sum_{cyc}\frac{a}{a^2+ab+b^2+3}\leq\sum_{cyc}\frac{a}{2(2a+\sqrt{ab})}=\frac{1}{2}\sum_{cyc}\frac{\sqrt{a}}{2\sqrt{a}+\sqrt{b}}\leq\frac{1}{2}.$$
But the inequality $a^2+ab+b^2+3\geq2(2a+\sqrt{ab})$ is wrong! Try $a=2$ and $b=\frac{1}{4}$
Also we can try to use a full expanding (I tried!) and to hope to use AM-GM,
but I think this way is very ugly and it's probably nothing.
Any hint would be desirable.
Thank you!

I think I get something :

Lemma

Let $f:\mathbb{R} \mapsto\mathbb{R^+}$ a real convex function for all $a,b,c,\alpha,\beta,\gamma$ positive real numbers then :
$$\frac{f(a)}{f(a)^2+f(a)f(b)+f(b)^2+3}+\frac{f(b)}{f(b)^2+f(b)f(c)+f(c)^2+3}+\frac{f(c)}{f(c)^2+f(c)f(a)+f(a)^2+3}\leq\frac{f(a)+\alpha f(b)}{\alpha(f(a)^2+f(a)f(b)+f(b)^2+3)}+\frac{f(b)+\beta f(c)}{\beta(f(b)^2+f(b)f(c)+f(c)^2+3)}+\frac{f(c)+\gamma f(a)}{\gamma(f(c)^2+f(c)f(a)+f(a)^2+3)}$$

Proof :

If we compare this :
$$\frac{f(a)}{f(a)^2+f(a)f(b)+f(b)^2+3}\leq \frac{f(a)+\alpha f(b)}{\alpha(f(a)^2+f(a)f(b)+f(b)^2+3)}\quad(1)$$
We get :
$$0\leq \alpha f(b)+(1-\alpha) f(a)$$
But with the definition of convexity we get :
$$0\leq f( (\alpha) (b)+(1-\alpha)(a))\leq\alpha f(b)+(1-\alpha) f(a)$$
So the inequality (1) is true , and sum to get the lemma .
Now put $\alpha=\frac{1}{2f(a)}$$\quad $$\beta=\frac{1}{2f(b)}$$\quad $$\gamma=\frac{1}{2f(c)}$ 
We get :
$$\frac{f(a)}{f(a)^2+f(a)f(b)+f(b)^2+3}+\frac{f(b)}{f(b)^2+f(b)f(c)+f(c)^2+3}+\frac{f(c)}{f(c)^2+f(c)f(a)+f(a)^2+3}\leq\frac{0.5+2f(a) f(b)}{(f(a)^2+f(a)f(b)+f(b)^2+3)}+\frac{0.5+f(b) f(c)}{(f(b)^2+f(b)f(c)+f(c)^2+3)}+\frac{0.5+f(c)f(a)}{(f(c)^2+f(c)f(a)+f(a)^2+3)}$$
But the minimum of :
$$\frac{0.5+2f(a) f(b)}{(f(a)^2+f(a)f(b)+f(b)^2+3)}$$ is reached when $f(a)=f(b)=0$
So we get :
$$\frac{f(a)}{f(a)^2+f(a)f(b)+f(b)^2+3}+\frac{f(b)}{f(b)^2+f(b)f(c)+f(c)^2+3}+\frac{f(c)}{f(c)^2+f(c)f(a)+f(a)^2+3}\leq 0.5$$
Done !
