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
\[\frac{t}{2} \cdot \frac{t}{3} + \frac{t}{2} \cdot \frac{t}{6} + \frac{t}{3} \cdot \frac{t}{6} = \frac{11}{36} t^2 = 11.\]
(x,y,z) = (x_0,y_0,z_0)일 때 등식이 발생한다고 가정합니다.$ 최소값을 찾고 증명하려면 다음과 같은 부등식을 조합해야 할 것 같습니다.
\[x^2 + y^2 \ge 2xy.\]$x = x_0$, $y = y_0$ 또는 $\frac{x}{x_0} = \frac{y}{y_0} = 1$일 때 부등식이 성립한다는 것을 기억하고 다음과 같이 부등식을 구성합니다.
\[\frac{x^2}{x_0^2} + \frac{y^2}{y_0^2} \ge \frac{2xy}{x_0 y_0}.\]그러면 다음과 같습니다.
\[\frac{y_0}{2x_0} \cdot x^2 + \frac{x_0}{2y_0} \cdot y^2 \ge xy.\]유사하게 계산합니다,
\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*}이를 더하면 다음과 같습니다.
\[\frac{y_0 + z_0}{2x_0} \cdot x^2 + \frac{x_0 + z_0}{2y_0} \cdot y^2 + \frac{x_0 + y_0}{2z_0} \cdot z^2 \ge xy + xz + yz.\]$x^2 + 2y^2 + 5z^2 = 22,$이므로 $x_0,$ $y_0,$ 및 $z_0$가 다음을 만족하기를 원합니다.
\[\frac{y_0 + z_0}{x_0} : \frac{x_0 + z_0}{y_0} : \frac{x_0 + y_0}{z_0} = 1:2:5.\]입니다.
\begin{align*}
Y_0 + Z_0 &= KX_0, \\.
x_0 + z_0 &= 2ky_0, \\.
X_0 + Y_0 &= 5KZ_0.
\end{align*}그렇다면
\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*}$t = x_0 + y_0 + z_0.$ 그러면 $x_0 = \frac{t}{k + 1},$ $y_0 = \frac{t}{2k + 1},$ 그리고 $z_0 = \frac{t}{5k + 1},$이므로 다음과 같이 됩니다.
\[\frac{t}{k + 1} + \frac{t}{2k + 1} + \frac{t}{5k + 1} = t.\]따라서,
\[\frac{1}{k + 1} + \frac{1}{2k + 1} + \frac{1}{5k + 1} = 1.\]이것은 $10k^3 - 8k - 2 = 0으로 단순화되며,$ 2(k - 1)(5k^2 + 5k + 1) = 0.$ $k$는 양수여야 하므로 $k = 1.$이 됩니다.

그런 다음 $x_0 = \frac{t}{2},$ $y_0 = \frac{t}{3},$ 및 $z_0 = \frac{t}{6}.$ 에 대입하면 $x^2 + 2y^2 + 5z^2 = 22,$ 를 얻습니다.
\[\frac{t^2}{4} + \frac{2t^2}{9} + \frac{5t^2}{36} = 22.\]풀면 $t = 6,$이고 $xy + xz + yz$의 최대값은 다음과 같습니다.
\[\frac{t}{2} \cdot \frac{t}{3} + \frac{t}{2} \cdot \frac{t}{6} + \frac{t}{3} \cdot \frac{t}{6} = \frac{11}{36} t^2 = 11.\]