137 Given that $x, y, z$ are all real numbers, and $x^{2}+y^{2}+z^{2}=x+y+z$, then the sum of the maximum and minimum values of $x+y+z$ is
A. 2
B. $2 \sqrt{2}$
C. 3
D. $2 \sqrt{3}$
137 C. Solution 1: It is clear that the minimum value of $x+y+z$ is 0. Also, $x+y+z+3=\left(x^{2}+1\right)+$ $\left(y^{2}+1\right)+\left(z^{2}+1\right) \geqslant 2 \cdot x+2 \cdot y+2 \cdot z$, from which we get $x+y+z \leqslant 3$. The equality holds when $x=$ $y=z=1$ and satisfies the given equation, so the maximum value of $x+y+z$ is 3.
Solution 2: By the Cauchy inequality, we have $3 \cdot(x+y+z)=\left(x^{2}+y^{2}+z^{2}\right) \cdot\left(1^{2}+1^{2}+\right.$ $\left.1^{2}\right) \geqslant(x \cdot 1+y \cdot 1+z \cdot 1)^{2}$. Since $x+y+z \geqslant 0$, we have $0 \leqslant x+y+z \leqslant$ 3, and both equalities can be achieved.
Solution 3: $x^{2}+y^{2}+z^{2}=x+y+z$ can be rewritten as $\left(x-\frac{1}{2}\right)^{2}+\left(y-\frac{1}{2}\right)^{2}+$ $\left(z-\frac{1}{2}\right)^{2}=\left(\frac{\sqrt{3}}{2}\right)^{2}$, which represents a sphere in the three-dimensional Cartesian coordinate system with center at $\left(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}\right)$ and radius $\frac{\sqrt{3}}{2}$. $x+y+z=x^{2}+y^{2}+z^{2}$ is the square of the distance from the point $(x, y, z)$ on this sphere to the origin, so the minimum value of $x^{2}+y^{2}+z^{2}$ is 0, and the maximum value is $\left(2 \cdot \frac{\sqrt{3}}{2}\right)^{2}=3$.