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

$$
(x-8)^{2}+(y-9)^{2}+(x-y)^{2}=\frac{1}{3}
$$
Answer: $x=8 \frac{1}{3}, y=8 \frac{2}{3}$

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

$$
\sqrt{\frac{(x-8)^{2}+(9-y)^{2}+(x-y)^{2}}{3}} \geqslant \frac{|x-8|+|9-y|+|x-y|}{3} \geqslant \frac{x-8+9-y+y-x}{3}=\frac{1}{3}
$$

That is, $(x-8)^{2}+(9-y)^{2}+(x-y)^{2} \geqslant \frac{1}{3}$. Equality is achieved only when $x-8=9-y=$ $y-x=\frac{1}{3}$. Then $x=8 \frac{1}{3}, y=8 \frac{2}{3}$.