Danube Mathematical Competition, 2010

Let $f(x_1, x_2, \dots, x_n) = (n-1)(x_1^2 + x_2^2 + \dots + x_n^2) + nx_1x_2 \dots x_n$. Assuming $0 \le x_1 \le x_2 \le \dots \le x_n$, we first show that
$$f(x_1, x_2, \dots, x_n) \ge f(x_1, (x_2+x_n)/2, x_3, \dots, x_{n-1}, (x_2+x_n)/2).$$

To begin with, notice that $x_n \ge 1$, to get
$$\begin{aligned} f(x_1, x_2, \dots, x_n) - f(x_1, (x_2+x_n)/2, x_3, \dots, x_{n-1}, (x_2+x_n)/2) \\ &= \frac{1}{4}(x_2 - x_n)^2 (2(n-1) - nx_1x_3 \cdots x_{n-1}) \\ &\ge \frac{1}{4}(x_2 - x_n)^2 (2(n-1) - nx_1x_3 \cdots x_{n-1}x_n). \end{aligned}$$
We now show that $x_1x_3\cdots x_{n-1}x_n \le 2(n-1)/n$, so the difference above is indeed non-negative. Notice first that
$$n = x_1 + x_2 + \dots + x_n \ge 2x_1 + x_2 + \dots + x_n \ge (n-1)(2x_1x_3\cdots x_{n-1}x_n)^{1/(n-1)},$$
so $x_1x_3\cdots x_{n-1}x_n \le n^{n-1}/(2(n-1)^{n-1})$. Since $n^{n-1}/(2(n-1)^{n-1}) \le 2(n-1)/n$, $n \ge 2$, the conclusion follows. (The latter follows from the well-known fact that the sequence $u_n = (1+1/n)^{n+1}$, $n \ge 1$, is decreasing, so $u_n \le u_1 = 4$, $n \ge 1$; that $u_n$ decreases is easily seen by Bernoulli's inequality: $n^{2(n+1)}/(n^2-1)^{n+1} = (1+1/(n^2-1))^{n+1} \ge 1 + (n+1)/(n^2-1) = n/(n-1)$, $n \ge 2$.)

Consequently, the value of $f$ does not increase when the largest and the second smallest of the $x_k$ is replaced by their arithmetic mean. It follows that $f$ achieves its minimum when $n-1$ of the $x_k$ are all equal to some $x$, and the smallest of the $x_k$ is $n - (n-1)x$. (That $f$ actually achieves a minimum subject to the stated conditions follows by a standard compactness argument which may be assumed without proof.) Since the $x_k$ are all non-negative, and $n - (n-1)x$ is the smallest among them, it follows that $1 \le x \le n/(n-1)$.
So we have to minimise the polynomial function
$$g(x) = (n-1)\left((n - (n-1)x)^2 + (n-1)x^2\right) + n(n - (n-1)x)x^{n-1}$$
on the closed interval $1 \le x \le n/(n-1)$. To this end, recall the guess made in the beginning, to evaluate
$$g(x) - n^2 = n(n - (n-1)x)(x-1)^2 \sum_{k=1}^{n-2} (x^{k-1} + x^{k-2} + \dots + 1) \ge 0,$$
$1 \le x \le n/(n-1)$. Equality holds if and only if $x=1$ or $x=n/(n-1)$, whence the conclusion.