10. Petya chooses non-negative numbers $x_{1}, x_{2}, \ldots, x_{11}$ such that their sum is 1. Vasya arranges them in a row as he sees fit, calculates the products of adjacent numbers, and writes the largest of the resulting ten products on the board. Petya wants the number on the board to be as large as possible, while Vasya wants it to be as small as possible. What number will end up on the board with the best play by Petya and Vasya? (A. Khryabrov)
Answer: $\frac{1}{40}$. Solution: If Petya chooses the numbers $\frac{1}{2}, \frac{1}{20}, \frac{1}{20}, \ldots, \frac{1}{20}$, then no matter how Vasya arranges these numbers, the adjacent numbers will be $\frac{1}{2}$ and $\frac{1}{20}$. Therefore, one of the products will be $\frac{1}{40}$, and the others will not exceed it. Thus, $\frac{1}{40}$ will appear on the board.

We will show how Vasya can obtain a number not greater than $\frac{1}{40}$ for any numbers. Let's renumber the numbers in descending order: $x_{1} \geq x_{2} \geq \ldots \geq x_{11}$. Arrange them in the following order: $x_{1}, x_{11}, x_{2}, x_{10}, x_{3}, x_{9}, x_{4}, x_{8}, x_{5}, x_{7}, x_{6}$. Then the products of adjacent numbers will be: $x_{1} x_{11} \geq x_{11} x_{2}, x_{2} x_{10} \geq x_{10} x_{3}, x_{3} x_{9} \geq x_{9} x_{4}, x_{4} x_{8} \geq x_{8} x_{5}, x_{5} x_{7} \geq x_{7} x_{6}$. We will show that they will not exceed $\frac{1}{40}$. For this, we will use two considerations: the arithmetic mean of several numbers is not less than the smallest of them and $x(1-x) \leq \frac{1}{4}$. It is sufficient to estimate $x_{1} x_{11}, x_{2} x_{10}, x_{3} x_{9}, x_{4} x_{8}$, and $x_{5} x_{7}$:

\[
\begin{aligned}
& x_{1} x_{11} \leq x_{1} \cdot \frac{x_{2}+x_{3}+\ldots+x_{11}}{10}=\frac{x_{1}\left(1-x_{1}\right)}{10} \leq \frac{1}{40} \\
& x_{2} x_{10} \leq \frac{x_{1}+x_{2}}{2} \cdot \frac{x_{3}+x_{4}+\ldots+x_{10}}{8} \leq \frac{\left(x_{1}+x_{2}\right)\left(1-\left(x_{1}+x_{2}\right)\right)}{16} \leq \frac{1}{64}<\frac{1}{40} \\
& x_{3} x_{9} \leq \frac{x_{1}+x_{2}+x_{3}}{3} \cdot \frac{x_{4}+x_{5}+\ldots+x_{9}}{6} \leq \frac{\left(x_{1}+x_{2}+x_{3}\right)\left(1-\left(x_{1}+x_{2}+x_{3}\right)\right)}{18} \leq \frac{1}{72}<\frac{1}{40} \\
& x_{4} x_{8} \leq \frac{x_{1}+x_{2}+x_{3}+x_{4}}{4} \cdot \frac{x_{5}+x_{6}+x_{7}+x_{8}}{4} \leq \frac{\left(x_{1}+x_{2}+x_{3}+x_{4}\right)\left(1-\left(x_{1}+x_{2}+x_{3}+x_{4}\right)\right)}{16} \leq \frac{1}{64}<\frac{1}{40} \\
& x_{5} x_{7} \leq \frac{x_{1}+x_{2}+\ldots+x_{5}}{5} \cdot \frac{x_{6}+x_{7}}{2} \leq \frac{\left(x_{1}+x_{2}+\ldots+x_{5}\right)\left(1-\left(x_{1}+x_{2}+\ldots+x_{5}\right)\right)}{10} \leq \frac{1}{40}
\end{aligned}
\]