, 2011

The required minimum is $n^2(n-1)$ and is achieved if and only if the $x_i$ are all equal to $n-1$.
Write
$$\sum_{i=1}^{n} x_i^2 \left( 1 + \frac{x_i^{n-2}}{x_1 \cdots x_{i-1} x_{i+1} \cdots x_n} \right) = \sum_{i=1}^{n} x_i^2 + \sum_{i=1}^{n} \frac{x_i^n}{x_1 \cdots x_{i-1} x_{i+1} \cdots x_n}$$
and minimise each sum separately.
To minimise the first sum, notice that
$$\sum_{i=1}^{n} x_i = \sum_{i=1}^{n} (x_i + 1) - n = \left( \sum_{i=1}^{n} (x_i + 1) \right) \sum_{i=1}^{n} \frac{1}{x_i + 1} - n \geq n^2 - n = n(n-1),$$
so
$$\sum_{i=1}^{n} x_i^2 \geq \frac{1}{n} \left( \sum_{i=1}^{n} x_i \right)^2 \geq n(n-1)^2;$$
clearly, equality holds if and only if the $x_i$ are all $n-1$.
To minimise the second sum, apply the AM-GM inequality to obtain
$$\frac{x_i^n}{x_1 \cdots x_{i-1} x_{i+1} \cdots x_n} + \sum_{j \neq i} x_j \geq n x_i, \quad i = 1, 2, \dots, n,$$
and sum over all $i$ to get
$$\sum_{i=1}^{n} \frac{x_i^n}{x_1 \cdots x_{i-1} x_{i+1} \cdots x_n} \geq \sum_{i=1}^{n} x_i \geq n(n-1);$$
again, equality holds if and only if the $x_i$ are all $n-1$. The conclusion follows.