Example 6 Given a positive integer $n \geqslant 2$, A and B play a game. A selects $2 n$ (possibly identical) non-negative numbers $x_{1}, x_{2}, \cdots, x_{2 n}$, whose sum equals 1. B then arranges them in a certain order on a circle, after which B calculates the product of each pair of adjacent numbers and writes the largest of the $2 n$ products on the blackboard. A hopes that the number written on the blackboard is as large as possible, while B hopes it is as small as possible. What will the number written on the blackboard be under the correct strategy? ${ }^{[5]}$
【Analysis】Answer: $\frac{1}{8(n-1)}$.
Assume without loss of generality that the numbers chosen by A are arranged in non-decreasing order as
$0 \leqslant x_{1} \leqslant x_{2} \leqslant \cdots \leqslant x_{2 n}$, where $\sum_{i=1}^{2 n} x_{i}=1$, and let $M=\max _{2 \leqslant k \leqslant n}\left\{x_{k} x_{2 n+2-k}\right\}$.
We will prove a lemma below.
Lemma: Regardless of how B places the numbers, the final number written on the blackboard is no less than $M$.
Proof: In fact, there exists $k^{\prime}\left(2 \leqslant k^{\prime} \leqslant n\right)$, such that
$$
\begin{array}{l}
M=x_{k^{\prime}} x_{2 n+2-k^{\prime}} \\
\text { Let } A_{k^{\prime}}=\left\{x_{2 n+2-k^{\prime}}, x_{2 n+3-k^{\prime}}, \cdots, x_{2 n}\right\} .
\end{array}
$$

If there are two numbers in $A_{k^{\prime}}$ that are adjacent on the circle, then their product is no less than $x_{2 n+2-k^{\prime}}^{2} \geqslant x_{k^{\prime}} x_{2 n+2-k^{\prime}}$; and if the numbers in $A_{k^{\prime}}$ are not adjacent on the circle, then there are at least $k^{\prime}$ numbers (since $\left|A_{k^{\prime}}\right|=$ $\left.k^{\prime}-1<n\right)$ that are adjacent to at least one number in $A_{k^{\prime}}$, among these $k^{\prime}$ numbers, there must be a number no less than $x_{k^{\prime}}$, the product of this number and the adjacent number in $A_{k^{\prime}}$ is no less than $x_{k^{\prime}} x_{2 n+2-k^{\prime}}=M$.
The lemma is proved.
If B places the $2 n$ numbers in the order
$$
x_{1}, x_{2 n}, x_{2}, x_{2 n-1}, \cdots, x_{n-1}, x_{n+2}, x_{n}, x_{n+1}
$$

along the circle in a counterclockwise direction, the maximum product of any two adjacent numbers can only be one of
$$
x_{2 n} x_{2}, x_{2 n-1} x_{3}, \cdots, x_{n+2} x_{n}
$$

Thus, the maximum product of adjacent numbers is exactly $M$.
In summary, under B's optimal strategy, the number written on the blackboard is $M$.
Therefore, the problem is equivalent to finding
$$
M=\max _{2 \leqslant k \leqslant n}\left\{x_{k} x_{2 n+2-k}\right\}
$$

On one hand, for any $k(2 \leqslant k \leqslant n)$, we have
$$
\begin{array}{l} 
x_{k} x_{2 n+2-k} \\
\leqslant \frac{x_{k}+x_{k+1}+\cdots+x_{2 n+1-k}}{2 n+2-2 k} . \\
\frac{x_{2 n+2-k}+x_{2 n+3-k}+\cdots+x_{2 n}}{k-1} \\
= \frac{1}{2(n+1-k)(k-1)} . \\
\left(x_{k}+x_{k+1}+\cdots+x_{2 n+1-k}\right) . \\
\left(x_{2 n+2-k}+x_{2 n+3-k}+\cdots+x_{2 n}\right) .
\end{array}
$$

By the AM-GM inequality, we get
$$
\begin{array}{l}
\left(x_{k}+x_{k+1}+\cdots+x_{2 n+1-k}\right) . \\
\left(x_{2 n+2-k}+x_{2 n+3-k}+\cdots+x_{2 n}\right) \\
\leqslant\left(\frac{x_{k}+x_{k+1}+\cdots+x_{2 n}}{2}\right)^{2} \leqslant \frac{1}{4} .
\end{array}
$$

Moreover, when $2 \leqslant k \leqslant n$, it is clear that
$$
(n+1-k)(k-1) \geqslant n-1 \text {. }
$$

Combining the above inequalities, we get
$$
x_{k} x_{2 n+2-k} \leqslant \frac{1}{8(n-1)}(2 \leqslant k \leqslant n),
$$

i.e., $M \leqslant \frac{1}{8(n-1)}$.
On the other hand, when A chooses
$$
x_{1}=0, x_{2}=\cdots=x_{2 n-1}=\frac{1}{4(n-1)}, x_{2 n}=\frac{1}{2}
$$

$M$ exactly reaches the maximum value $\frac{1}{8(n-1)}$, which is A's optimal strategy.

Therefore, under the optimal strategies of A and B, the final number written on the blackboard is $\frac{1}{8(n-1)}$.