5. Let $x, y, z \in \mathbf{R}^{+}$, satisfying $x+y+z=x y z$, then the minimum value of the function $f(x, y, z)=x^{2}(y z-1)+y^{2}(z x-1)+z^{2}(x y-1)$ is $\qquad$ .
Answer 18.
Analysis According to the conditions, we have $y+z=x(y z-1)$, so $y z-1=\frac{y+z}{x}$. Similarly, we have $z x-1=\frac{z+x}{y}, x y=1=\frac{x+y}{z}$. Also, since $x y z=x+y+z \geq 3 \sqrt[3]{x y z}$, we get $x y z \geq 3 \sqrt{3}$, thus
$$
f(x, y, z)=2(x y+y z+z x) \geq 2 \cdot 3 \sqrt[3]{(x y z)^{2}} \geq 18
$$

Equality holds when $x=y=z=\sqrt{3}$.