Since $x$, $y$, $z$ are positive integers with sum $12$, it is enough to prove that
$$\frac{x}{y} + \frac{y}{z} + \frac{z}{x} + \frac{x+y+z}{4} \geq \sqrt{x} + \sqrt{y} + \sqrt{z}. \quad (1)$$
From the inequality of the arithmetic–geometric mean for the positive integers $x$, $y$, $z$ we get

$$\frac{x}{y} + \frac{y}{4} \ge 2 \sqrt{\frac{x}{y} \cdot \frac{y}{4}} = \sqrt{x}, \qquad (2)$$
$$\frac{y}{z} + \frac{z}{4} \ge 2 \sqrt{\frac{y}{z} \cdot \frac{z}{4}} = \sqrt{y}, \qquad (3)$$
$$\frac{z}{x} + \frac{x}{4} \ge 2 \sqrt{\frac{z}{x} \cdot \frac{x}{4}} = \sqrt{z}. \qquad (4)$$
Summing up (2), (3) and (4) we find inequality (1).

Equality holds when all inequalities (2), (3) and (4) hold as equalities, that is when
$$\begin{align*} \frac{x}{y} = \frac{y}{4}, \quad \frac{y}{z} = \frac{z}{4}, \quad \frac{z}{x} = \frac{x}{4} &\Leftrightarrow x = \frac{y^2}{4}, \quad y = \frac{z^2}{4}, \quad z = \frac{1}{4} \left(\frac{y^2}{4}\right)^2 = \frac{y^4}{4^3} \\ &\Leftrightarrow x = \frac{y^2}{4}, \quad y = \frac{z^2}{4}, \quad z = \frac{1}{4^3} \left(\frac{z^2}{4}\right)^4 \\ &\Leftrightarrow x = \frac{y^2}{4}, \quad y = \frac{z^2}{4}, \quad z = \frac{1}{4^7} z^8 \\ &\Leftrightarrow x = \frac{y^2}{4}, \quad y = \frac{z^2}{4}, \quad z^7 = 4^7 \\ &\Leftrightarrow x = y = z = 4. \end{align*}$$