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.

The idea is indeed that of using Cauchy-Schwarz. The problem is you went from a homogeneous expression ($a^2+4b^2+c^2$) to non-homogeneous terms and that may be harder to prove than the original inequality.
Instead, I would apply Cauchy-Schwarz like this:
$$\frac{(a+b+c)^2}{a^2+4b^2+c^2} \leq \frac{a^2}{a^2+b^2}+\frac{b^2}{2b^2}+\frac{c^2}{b^2+c^2}=\frac{a^2}{a^2+b^2}+\frac{c^2}{b^2+c^2}+\frac{1}{2}$$
and summing cyclically:
$$
\begin{aligned}
\sum \frac{(a+b+c)^2}{a^2+4b^2+c^2} &\leq \sum\left(\frac{a^2}{a^2+b^2}+\frac{c^2}{b^2+c^2}\right)+\frac{3}{2}\\
&=\sum\left(\frac{a^2}{a^2+b^2}+\frac{b^2}{a^2+b^2}\right)+\frac{3}{2}\\
&=3+\frac{3}{2}=\frac{9}{2}
\end{aligned}
$$
