, 2011

The required maximum is $1/(n-1)^n$ and is achieved if and only if the $x_i$ are all equal to $1/(n-1)$.
The constraint on the $x_i$ is equivalent to
$$\sum_{k=1}^{n} (k-1)\sigma_k = 1,$$
where
$$\sigma_k = \sum_{1 \le i_1 < \dots < i_k \le n} x_{i_1} \cdots x_{i_k}, \quad k = 1, 2, \dots, n.$$
By the AM-GM inequality,
$$\sigma_k \ge \binom{n}{k} \sigma_n^{k/n}, \quad k = 1, 2, \dots, n,$$
so, upon substitution $t = \sigma_n^{1/n}$,
$$\sum_{k=1}^{n} (k-1) \binom{n}{k} t^k \le 1;$$
that is, $(t+1)^{n-1}((n-1)t-1) \le 0$. Consequently, $t \le 1/(n-1)$. Equality clearly forces all the $x_i$ to be equal to $1/(n-1)$. Since these $x_i$ obey the constraint, the conclusion follows.