Given positives $a, b, c$ such that $a + b + c = 3$, prove that $\sum_{cyc}\frac{1}{a^2 + 4b^2 + c^2} \le \frac{1}{2}$. 
Given positives $a, b, c$ such that $a + b + c = 3$, prove that $$\frac{1}{c^2 + 4a^2 + b^2} + \frac{1}{a^2 + 4b^2 + c^2} + \frac{1}{b^2 + 4c^2 + a^2} \le \frac{1}{2}$$

We have that $$a^2 + 4b^2 + c^2 = a^2 + (a + b + c + 1)b^2 + c^2 = (b^2 + a)a + (b + 1)b^2 + (b^2 + c)c$$
$$\implies \sum_{cyc}\frac{1}{a^2 + 4b^2 + c^2} = \frac{1}{(a + b + c)^2} \cdot \sum_{cyc}\frac{(a + b + c)^2}{(b^2 + a)a + (b + 1)b^2 + (b^2 + c)c}$$
$$ \le \frac{1}{(a + b + c)^2} \cdot \sum_{cyc}\left(\frac{a}{b^2 + a} + \frac{1}{b + 1} + \frac{c}{b^2 + c}\right)$$
Furthermore, $\dfrac{a}{c^2 + a} + \dfrac{c}{a^2 + c} = 1 - \dfrac{(c + a - 2)ca}{(c^2 + a)(a^2 + c)}$, $$\sum_{cyc}\frac{1}{a^2 + 4b^2 + c^2} \le \frac{1}{(a + b + c)^2} \cdot \sum_{cyc}\left[\frac{1}{b + 1} - \frac{(c + a - 2)ca}{(c^2 + a)(a^2 + c)}\right] + \frac{1}{3}$$
Then I don't know what to do next.

We need to prove that
$$\sum_{cyc}\frac{1}{b^2+c^2+4a^2}\leq\frac{9}{2(a+b+c)^2}$$ or
$$\sum_{cyc}(2a^6-4a^5b-4a^5c+13a^4b^2+13a^4c^2-4a^4bc-12a^3b^3-12a^3b^2c-12a^3c^2b+20a^2b^2c^2)\geq0$$ or
$$\sum_{cyc}(a-b)^2(2c^4+2(a^2-4ab+b^2)c^2+a^4-2a^3b+4a^2b^2-2ab^3+b^4)\geq0$$ for which it's enough to prove that:
$$(a^2-4ab+b^2)^2-2(a^4-2a^3b+4a^2b^2-2ab^3+b^4)\leq0$$ or $$(a-b)^2(a^2+6ab+b^2)\geq0$$ and we are done!
