Solution. To prove the second inequality, we may assume $x = \max\{x, y, z\}$. Then we get
$$\frac{4z^2 + x^2}{z^2 + 4x^2} \le 1, \quad \frac{4x^2 + y^2}{x^2 + 4y^2} < 4, \quad \text{and} \quad \frac{4y^2 + z^2}{y^2 + 4z^2} < 4.$$
So the desired inequality follows.

Next we prove the first inequality. By the AM-GM inequality, we have $4xy^2 \le y^3 + 4x^2y$. Then
$$\begin{aligned} y^3 + 4x^2y + 3x^3 &> y^3 + 4x^2y \ge 4xy^2 > 3xy^2 \\ \iff y^3 + 4x^2y + 4x^3 + xy^2 &> 4xy^2 + x^3 \\ \iff \frac{4x^2 + y^2}{x^2 + 4y^2} &> \frac{x}{x+y}. \end{aligned}$$
Hence
$$\sum_{cyc} \frac{x}{x+y} < \sum_{cyc} \frac{4x^2 + y^2}{x^2 + 4y^2}.$$
Now apply the Cauchy-Schwartz inequality, we have
$$\sum_{cyc} \frac{x}{\sqrt{2(x^2 + y^2)}} = \sum_{cyc} \frac{x}{\sqrt{(1+1)(x^2 + y^2)}} \le \sum_{cyc} \frac{x}{x+y} < \sum_{cyc} \frac{4x^2 + y^2}{x^2 + 4y^2},$$
which is the first inequality. $\square$