9. (20 points) Find all values of $x$ and $y$ for which the following equality holds:

$$
(x-6)^{2}+(y-7)^{2}+(x-y)^{2}=\frac{1}{3}
$$
Answer: $x=6 \frac{1}{3}, y=6 \frac{2}{3}$

Solution: We use the inequality between the quadratic mean and the arithmetic mean:

$$
\sqrt{\frac{(x-6)^{2}+(7-y)^{2}+(x-y)^{2}}{3}} \geqslant \frac{|x-6|+|7-y|+|x-y|}{3} \geqslant \frac{x-6+7-y+y-x}{3}=\frac{1}{3}
$$

That is, $(x-6)^{2}+(7-y)^{2}+(x-y)^{2} \geqslant \frac{1}{3}$. Equality is achieved only when $x-6=7-y=$ $y-x=\frac{1}{3}$. Then $x=6 \frac{1}{3}, y=6 \frac{2}{3}$.