By QM-AM, \[\sqrt{\frac{x^2 + y^2}{2}} \ge \frac{x + y}{2}.\]Then $ \sqrt{x^2 + y^2} \ge \frac{x + y}{\sqrt{2}} $. Similarly, \begin{align*} \sqrt{x^2 + z^2} &\ge \frac{x + z}{\sqrt{2}}, \\ \sqrt{y^2 + z^2} &\ge \frac{y + z}{\sqrt{2}}, \end{align*}so \[\sqrt{x^2 + y^2} + \sqrt{x^2 + z^2} + \sqrt{y^2 + z^2} \ge \frac{x + y}{\sqrt{2}} + \frac{x + z}{\sqrt{2}} + \frac{y + z}{\sqrt{2}} = \sqrt{2} (x + y + z).\]Therefore, \[\frac{\sqrt{x^2 + y^2} + \sqrt{x^2 + z^2} + \sqrt{y^2 + z^2}}{x + y + z} \ge \sqrt{2}.\]Equality occurs when $ x = y = z, $ so the minimum value is $ \boxed{\sqrt{2}} $.