Suppose equality occurs when $ (x,y,z) = (x_0,y_0,z_0) $. To find and prove the minimum value, it looks like we're going to have to put together some inequalities like \[x^2 + y^2 \ge 2xy.\]Remembering that equality occurs when $ x = x_0 $ and $ y = y_0, $ or $ \frac{x}{x_0} = \frac{y}{y_0} = 1, $ we form the inequality \[\frac{x^2}{x_0^2} + \frac{y^2}{y_0^2} \ge \frac{2xy}{x_0 y_0}.\]Then \[\frac{y_0}{2x_0} \cdot x^2 + \frac{x_0}{2y_0} \cdot y^2 \ge xy.\]Similarly, \begin{align*} \frac{z_0}{2x_0} \cdot x^2 + \frac{x_0}{2z_0} \cdot z^2 \ge xz, \\ \frac{z_0}{2y_0} \cdot y^2 + \frac{y_0}{2z_0} \cdot z^2 \ge xz.\end{align*}Adding these, we get \[\frac{y_0 + z_0}{2x_0} \cdot x^2 + \frac{x_0 + z_0}{2y_0} \cdot y^2 + \frac{x_0 + y_0}{2z_0} \cdot z^2 \ge xy + xz + yz.\]Since we are given that $ x^2 + 2y^2 + 5z^2 = 22, $ we want $ x_0, $ $ y_0, $ and $ z_0 $ to satisfy \[\frac{y_0 + z_0}{x_0} : \frac{x_0 + z_0}{y_0} : \frac{x_0 + y_0}{z_0} = 1:2:5.\]Let \begin{align*} y_0 + z_0 &= kx_0, \\ x_0 + z_0 &= 2ky_0, \\ x_0 + y_0 &= 5kz_0.\end{align*}Then \begin{align*} x_0 + y_0 + z_0 &= (k + 1) x_0, \\ x_0 + y_0 + z_0 &= (2k + 1) y_0, \\ x_0 + y_0 + z_0 &= (5k + 1) z_0.\end{align*}Let $ t = x_0 + y_0 + z_0 $. Then $ x_0 = \frac{t}{k + 1}, $ $ y_0 = \frac{t}{2k + 1}, $ and $ z_0 = \frac{t}{5k + 1}, $ so \[\frac{t}{k + 1} + \frac{t}{2k + 1} + \frac{t}{5k + 1} = t.\]Hence, \[\frac{1}{k + 1} + \frac{1}{2k + 1} + \frac{1}{5k + 1} = 1.\]This simplifies to $ 10k^3 - 8k - 2 = 0, $ which factors as $ 2(k - 1)(5k^2 + 5k + 1) = 0 $. Since $ k $ must be positive, $ k = 1 $. Then $ x_0 = \frac{t}{2}, $ $ y_0 = \frac{t}{3}, $ and $ z_0 = \frac{t}{6} $. Substituting into $ x^2 + 2y^2 + 5z^2 = 22, $ we get \[\frac{t^2}{4} + \frac{2t^2}{9} + \frac{5t^2}{36} = 22.\]Solving, we find $ t = 6, $ and the maximum value of $ xy + xz + yz $ is \[\frac{t}{2} \cdot \frac{t}{3} + \frac{t}{2} \cdot \frac{t}{6} + \frac{t}{3} \cdot \frac{t}{6} = \frac{11}{36} t^2 = \boxed{11}.\]