jmc

By QM-AM, $$\sqrt{\frac{x^2+y^2}{2}}\ge \frac{x+y}{2}.$$Then $$\frac{x^2+y^2}{2}\ge {\left(\frac{x+y}{2}\right)}^2,$$which we can re-arrange as $$\frac{x^2+y^2}{x+y}\ge \frac{x+y}{2}.$$Similarly, $$\begin{array}{rl}\frac{x^2+y^2}{x+y}& \ge \frac{x+y}{2},\\ \frac{y^2+z^2}{y+z}& \ge \frac{y+z}{2}.\end{array}$$Therefore, $$\frac{x^2+y^2}{x+y}+\frac{x^2+z^2}{x+z}+\frac{y^2+z^2}{y+z}\ge \frac{x+y}{2}+\frac{x+z}{2}+\frac{y+z}{2}=x+y+z=6.$$Equality occurs when $x=y=z=2,$ so the minimum value is $\boxed{6}.$