jmc

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, $$\begin{array}{rl}\sqrt{x^2+z^2}& \ge \frac{x+z}{\sqrt{2}},\\ \sqrt{y^2+z^2}& \ge \frac{y+z}{\sqrt{2}},\end{array}$$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 $\boxed{\sqrt{2}}.$