9.10. Petya and Vasya are playing the following game. Petya chooses 100 (not necessarily distinct) non-negative numbers $x_{1}, x_{2}, \ldots$, $x_{100}$, the sum of which is 1. Vasya then pairs them up as he sees fit, calculates the product of the numbers in each pair, and writes the largest of the 50 obtained 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 be on the board with the correct play?

(A. Khryabrov)
Answer: $\frac{1}{396}$.

Solution: If Petya chooses the numbers $\frac{1}{2}, \frac{1}{198}, \frac{1}{198}, \ldots, \frac{1}{198}$, then no matter how Vasya pairs these numbers, the number $\frac{1}{198}$ will be paired with $\frac{1}{2}$. Their product will be $\frac{1}{396}$, and the other products will not exceed it. Therefore, the number $\frac{1}{396}$ will appear on the board.

We will show how Vasya can obtain a number not greater than $\frac{1}{396}$ for any set of Petya's numbers. Let's renumber the numbers in non-increasing order: $x_{1} \geqslant x_{2} \geqslant \ldots \geqslant x_{100}$. We will pair the numbers as follows: $x_{k}$ with $x_{101-k}$. The products of the numbers in the pairs will be

$$
x_{1} x_{100}, x_{2} x_{99}, x_{3} x_{98}, \ldots, x_{k} x_{101-k}, \ldots, x_{50} x_{51}
$$

We will show that $a=x_{k} x_{101-k} \leqslant \frac{1}{396}$ for $k \leqslant 49$. Indeed, from the inequalities $x_{k} \leqslant x_{k-1} \leqslant \ldots \leqslant x_{1}$, it follows that $k x_{k} \leqslant x_{1}+x_{2}+\ldots+x_{k}$, so

$$
k a=k x_{k} \cdot x_{101-k} \leqslant\left(x_{1}+x_{2}+\ldots+x_{k}\right) x_{101-k}
$$

Similarly, from the inequalities $x_{101-k} \leqslant x_{100-k} \leqslant x_{99-k} \leqslant \ldots \leqslant x_{k+1}$, it follows that

$$
\begin{aligned}
& (101-2 k) x_{101-k} \leqslant x_{101-k}+x_{100-k}+\ldots+x_{k+1} \leqslant \\
& \quad \leqslant x_{k+1}+x_{k+2}+\ldots+x_{100}=1-x_{1}-x_{2}-\ldots-x_{k}
\end{aligned}
$$

Therefore,

$k(101-2 k) a \leqslant\left(x_{1}+x_{2}+\ldots+x_{k}\right)\left(1-x_{1}-x_{2}-\ldots-x_{k}\right)=x(1-x)$, where $x=x_{1}+x_{2}+\ldots+x_{k}$. Since by the inequality of means for two numbers $x(1-x) \leqslant\left(\frac{x+(1-x)}{2}\right)^{2}=\frac{1}{4}$, we get the inequality $x_{k} x_{101-k}=a \leqslant \frac{1}{4 k(101-2 k)}$. It remains to prove that $k(101-2 k) \geqslant 99$ for $k \leqslant 49$. This inequality can be rewritten as $(k-1)(99-2 k) \geqslant 0$, and both brackets in the last formula are non-negative.

It remains to prove that $x_{50} x_{51} \leqslant \frac{1}{396}$. Since $x_{50} \leqslant x_{49} \leqslant x_{48} \leqslant \ldots \leqslant x_{2} \leqslant x_{1}$, we have

$$
x_{50} \leqslant \frac{x_{1}+x_{2}+\ldots+x_{50}}{50} \leqslant \frac{1}{50}
$$

and, similarly,

$$
x_{51} \leqslant \frac{x_{1}+x_{2}+\ldots+x_{51}}{51} \leqslant \frac{1}{51}
$$

Therefore, $x_{50} x_{51} \leqslant \frac{1}{50 \cdot 51}<\frac{1}{396}$.

Comment: Only the answer - 0 points.

Only an example of Petya's numbers, for which $\frac{1}{396}$ will appear on the board - 1 point.

Only a proof that Vasya can always obtain a number not less than $\frac{1}{396}$ - 5 points.

Only indicating that Vasya should pair the numbers $x_{1} \geqslant \geqslant \ldots \geqslant x_{101}$ in pairs of the form $x_{k} x_{101-k}$ does not add points.

If the work proves that with such pairing, the number on the board will not be greater than with any other pairing - 1 point (this point can be combined with the point for the example).