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
\[\frac{y_0 + z_0}{2x_0} \cdot x^2 + \frac{x_0 + z_0}{2y_0} \cdot y^2 + \frac{x_0 + y_0}{2z_0} \cdot z^2 \ge xy + xz + yz.\]We want to maximize $10x^2 + 10y^2 + z^2,$ so we want $x_0,$ $y_0,$ and $z_0$ to satisfy
\[\frac{y_0 + z_0}{x_0} : \frac{x_0 + z_0}{y_0} : \frac{x_0 + y_0}{z_0} = 10:10:1.\]Let
\begin{align*}
y_0 + z_0 &= 10kx_0, \\
x_0 + z_0 &= 10ky_0, \\
x_0 + y_0 &= kz_0.
\end{align*}Then
\begin{align*}
x_0 + y_0 + z_0 &= (10k + 1) x_0, \\
x_0 + y_0 + z_0 &= (10k + 1) y_0, \\
x_0 + y_0 + z_0 &= (k + 1) z_0.
\end{align*}Let $t = x_0 + y_0 + z_0.$  Then $x_0 = \frac{t}{10k + 1},$ $y_0 = \frac{t}{10k + 1},$ and $z_0 = \frac{t}{k + 1},$ so
\[\frac{t}{10k + 1} + \frac{t}{10k + 1} + \frac{t}{k + 1} = t.\]Hence,
\[\frac{1}{10k + 1} + \frac{1}{10k + 1} + \frac{1}{k + 1} = 1.\]This simplifies to $10k^2 - k - 2 = 0,$ which factors as $(2k - 1)(5k + 2) = 0.$  Since $k$ is positive, $k = \frac{1}{2}.$

Then $x_0 = \frac{t}{6},$ $y_0 = \frac{t}{6},$ and $z_0 = \frac{2t}{3}.$  Substituting into $xy + xz + yz = 1,$ we get
\[\frac{t^2}{36} + \frac{t^2}{9} + \frac{t^2}{9} = 1.\]Solving, we find $t = 2,$ and the minimum value of $10x^2 + 10y^2 + z^2$ is
\[10 \cdot \frac{t^2}{36} + 10 \cdot \frac{t^2}{36} + \frac{4t^2}{9} = t^2 = 4.\]
(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.
\[\frac{y_0 + z_0}{2x_0} \cdot x^2 + \frac{x_0 + z_0}{2y_0} \cdot y^2 + \frac{x_0 + y_0}{2z_0} \cdot z^2 \ge xy + xz + yz.\]$10x^2 + 10y^2 + z^2,$를 최대화하려면 $x_0,$ $y_0,$ 및 $z_0$이 만족되기를 원합니다.
\[\frac{y_0 + z_0}{x_0} : \frac{x_0 + z_0}{y_0} : \frac{x_0 + y_0}{z_0} = 10:10:1.\]입니다.
\begin{align*}
Y_0 + Z_0 &= 10KX_0, \\.
x_0 + z_0 &= 10ky_0, \\.
X_0 + Y_0 &= KZ_0.
\end{align*}그렇다면
\begin{align*}
x_0 + y_0 + z_0 &= (10k + 1) x_0, \\.
x_0 + y_0 + z_0 &= (10k + 1) y_0, \\
X_0 + Y_0 + Z_0 &= (k + 1) Z_0.
\end{align*}$t = x_0 + y_0 + z_0.$ 그러면 $x_0 = \frac{t}{10k + 1},$ $y_0 = \frac{t}{10k + 1},$ 그리고 $z_0 = \frac{t}{k + 1},$이므로 다음과 같이 됩니다.
\[\frac{t}{10k + 1} + \frac{t}{10k + 1} + \frac{t}{k + 1} = t.\]따라서,
\[\frac{1}{10k + 1} + \frac{1}{10k + 1} + \frac{1}{k + 1} = 1.\]이것은 $10k^2 - k - 2 = 0으로 단순화되며,$(2k - 1)(5k + 2) = 0.$ $k$는 양수이므로 $k = \frac{1}{2}.$로 인수분해합니다.

그러면 $x_0 = \frac{t}{6},$ $y_0 = \frac{t}{6},$ 및 $z_0 = \frac{2t}{3}.$ 에 대입하면 $xy + xz + yz = 1,$ 을 얻습니다.
\[\frac{t^2}{36} + \frac{t^2}{9} + \frac{t^2}{9} = 1.\]풀면 $t = 2,$이고 최소값 $10x^2 + 10y^2 + z^2$ 는 다음과 같습니다.
\[10 \cdot \frac{t^2}{36} + 10 \cdot \frac{t^2}{36} + \frac{4t^2}{9} = t^2 = 4.\]입니다.