First note that the function $x \mapsto x^{-\frac{1}{2}}$ is convex on $(0, \infty)$. Thus by Jensen's inequality, we have
$$\begin{aligned} & \frac{x}{x+y+z} (2y+3z)^{-\frac{1}{2}} + \frac{y}{x+y+z} (2z+3x)^{-\frac{1}{2}} + \frac{z}{x+y+z} (2x+3y)^{-\frac{1}{2}} \\ \ge & \left( \frac{x(2y+3z) + y(2z+3x) + z(2x+3y)}{x+y+z} \right)^{-\frac{1}{2}} \\ = & \frac{1}{\sqrt{5}} \cdot \sqrt{\frac{x+y+z}{xy+yz+zx}} \end{aligned}$$
Hence it suffices to show that
$$x+y+z \ge xy+yz+zx. \tag{1}$$
Without loss of generality, we may assume that $x \le y \le z$. Then from the condition $x+y+z+xyz=4$, we see that $x \le 1 \le z$. Thus $(z-1)(1-x) = x+z-1-zx \ge 0$. It follows that
$$(1+zx)(x+y+z-zx) = (x+y+z+xyz) + zx(x+z-1-zx) \ge 4. \tag{2}$$
On the other hand, by the AM-GM inequality, we have
$$4 \ge (4-(x+z))(x+z) = (y+xyz)(x+z) = (1+zx)(xy+yz). \tag{3}$$
Combining (2) and (3), we get (1), since $1+zx > 0$.