shortlistBMO 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 $$\sum _{i=1}^n{x}_i=\sum _{i=1}^n(x_i+1)-n=\left(\sum _{i=1}^n(x_i+1)\right)\sum _{i=1}^n\frac{1}{x_i+1}-n\ge n^2-n=n(n-1),$$ so $$\sum _{i=1}^n{x}_i^2\ge \frac{1}{n}{\left(\sum _{i=1}^n{x}_i\right)}^2\ge n(n-1{)}^2;$$ 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 $$\frac{x_i^n}{x_1\cdots x_{i-1}{x}_{i+1}\cdots x_n}+\sum _{j\ne i}{x}_j\ge n{x}_i,i=1,2,\dots ,n,$$ and sum over all $i$ to get $$\sum _{i=1}^n\frac{x_i^n}{x_1\cdots x_{i-1}{x}_{i+1}\cdots x_n}\ge \sum _{i=1}^n{x}_i\ge n(n-1);$$ again, equality holds if and only if the $x_i$ are all $n-1$. The conclusion follows.