Our strategy is to take $ x^2 + y^2 + z^2 $ and divide into several expression, apply AM-GM to each expression, and come up with a multiple of $ xy \sqrt{10} + yz $. Since we want terms of $ xy $ and $ yz $ after applying AM-GM, we divide $ x^2 + y^2 + z^2 $ into \[(x^2 + ky^2) + [(1 - k)y^2 + z^2].\]By AM-GM, \begin{align*} x^2 + ky^2 &\ge 2 \sqrt{(x^2)(ky^2)} = 2xy \sqrt{k}, \\ (1 - k)y^2 + z^2 &\ge 2 \sqrt{((1 - k)y^2)(z^2)} = 2yz \sqrt{1 - k}.\end{align*}To get a multiple of $ xy \sqrt{10} + yz, $ we want $ k $ so that \[\frac{2 \sqrt{k}}{\sqrt{10}} = 2 \sqrt{1 - k}.\]Then \[\frac{\sqrt{k}}{\sqrt{10}} = \sqrt{1 - k}.\]Squaring both sides, we get \[\frac{k}{10} = 1 - k.\]Solving for $ k, $ we find $ k = \frac{10}{11} $. Thus, \begin{align*} x^2 + \frac{10}{11} y^2 &\ge 2xy \sqrt{\frac{10}{11}}, \\ \frac{1}{11} y^2 + z^2 &\ge 2yz \sqrt{\frac{1}{11}}, \end{align*}so \[1 = x^2 + y^2 + z^2 \ge 2xy \sqrt{\frac{10}{11}} + 2yz \sqrt{\frac{1}{11}}.\]Multiplying by $ \sqrt{11}, $ we get \[2xy \sqrt{10} + 2yz \le \sqrt{11}.\]Dividing by 2, we get \[xy \sqrt{10} + yz \le \frac{\sqrt{11}}{2}.\]Equality occurs when $ x = y \sqrt{\frac{10}{11}} $ and $ y \sqrt{\frac{1}{11}} = z $. Using the condition $ x^2 + y^2 + z^2 = 1, $ we can solve to get $ x = \sqrt{\frac{10}{22}}, $ $ y = \sqrt{\frac{11}{22}}, $ and $ z = \sqrt{\frac{1}{22}}, $ so the minimum value is $ \boxed{\frac{\sqrt{11}}{2}} $.