jmc

By Pythagoras, the lengths $x,$ $\sqrt{16-x^2},$ and 4 are the sides of a right triangle. Similarly, $y,$ $\sqrt{25-y^2},$ and 5 are the sides of a right triangle, and $z,$ $\sqrt{36-z^2},$ and 6 are the sides of a right triangle. Stack these right triangles, as shown below. Then $AE=x+y+z=9$ and $$DE=\sqrt{16-x^2}+\sqrt{25-y^2}+\sqrt{36-z^2}.$$

By the Triangle Inequality, $$AD\le AB+BC+CD=4+5+6=15.$$By Pythagoras on right triangle $ADE,$ $$9^2+(\sqrt{16-x^2}+\sqrt{25-y^2}+\sqrt{36-z^2}{)}^2=A{D}^2\le 15^2,$$so $(\sqrt{16-x^2}+\sqrt{25-y^2}+\sqrt{36-z^2}{)}^2\le 15^2-9^2=144.$ Hence, $$\sqrt{16-x^2}+\sqrt{25-y^2}+\sqrt{36-z^2}\le 12.$$Equality occurs when $x=\frac{12}{5},$ $y=3,$ and $z=\frac{18}{5}.$ (Note that this corresponds to the case where $A,$ $B,$ $C,$ and $D$ are collinear.) Thus, the maximum value we seek is $\boxed{12}.$