Solution:
Equality holds when $x = y = z = 0$.
Apply AM-GM to $x + y + z = x y z$,
$$\begin{aligned} & x y z = x + y + z \geq 3 \sqrt[3]{x y z} \Rightarrow (x y z)^{3} \geq (3 \sqrt[3]{x y z})^{3} \\ & \Rightarrow x^{3} y^{3} z^{3} \geq 27 x y z \\ & \Rightarrow x^{2} y^{2} z^{2} \geq 27 \\ & \Rightarrow \sqrt[3]{x^{2} y^{2} z^{2}} \geq 3 \end{aligned}$$
Also by AM-GM we have, $x^{2} + y^{2} + z^{2} \geq 3 \sqrt[3]{x^{2} y^{2} z^{2}} \geq 9$.
Therefore we get $x^{2} + y^{2} + z^{2} \geq 9$.
Now,
$$\begin{gathered} 2\left(x^{2} + y^{2} + z^{2}\right) \geq 3(x + y + z) \Leftrightarrow \frac{2\left(x^{2} + y^{2} + z^{2}\right)}{3} \geq (x + y + z) \\ \Leftrightarrow 2 \cdot \frac{2\left(x^{2} + y^{2} + z^{2}\right)}{3} \geq 2 \cdot (x + y + z) \\ \Leftrightarrow \frac{4\left(x^{2} + y^{2} + z^{2}\right)}{3} \geq 2 \cdot (x + y + z) \\ \Leftrightarrow x^{2} + y^{2} + z^{2} + \frac{\left(x^{2} + y^{2} + z^{2}\right)}{3} \geq 2 \cdot (x + y + z) \\ \Leftrightarrow 3 + \frac{x^{2}}{3} + 3 + \frac{y^{2}}{3} + 3 + \frac{z^{2}}{3} \geq 2(x + y + z) \\ \Leftrightarrow 3 + \frac{x^{2}}{3} + 3 + \frac{y^{2}}{3} + 3 + \frac{z^{2}}{3} \geq 2(x + y + z) \\ \Leftrightarrow 2 \sqrt{3 \cdot \frac{x^{2}}{3}} + 2 \sqrt{3 \cdot \frac{y^{2}}{3}} + 2 \sqrt{3 \cdot \frac{z^{2}}{3}} \geq 2(x + y + z) \end{gathered}$$
Equality holds if $3 = \frac{x^{2}}{3} = \frac{y^{2}}{3} = \frac{z^{2}}{3}$, i.e. $x = y = z = 3$, for which $x + y + z \neq x y z$.

Solution:
If one of the numbers is zero, then from $x + y + z = x y z$ all three numbers are zero and the equality trivially holds.
From AM-GM $x^{2} + y^{2} + z^{2} \geq 3 \sqrt[3]{x^{2} y^{2} z^{2}} = 3 \frac{x + y + z}{\sqrt[3]{x y z}} \geq 3 \frac{x + y + z}{\frac{x + y + z}{3}} = 9$
From QM-AM $\frac{x^{2} + y^{2} + z^{2}}{3} \geq \left(\frac{x + y + z}{3}\right)^{2}$
Multiplying (1) and (2) we get $\frac{\left(x^{2} + y^{2} + z^{2}\right)^{2}}{3} \geq 9 \frac{(x + y + z)^{2}}{9} = (x + y + z)^{2}$. By taking square root on both sides we deduce the stated inequality.
Equality holds only when $x = y = z = \sqrt{3}$ or $x = y = z = 0$.