Romania — NMO Selection Tests for the Balkan and International Mathematical Olympiads

By the power-mean inequality,
$$1 + a^n \geq \frac{(1 + a)^n}{2^{n-1}}, \quad a \geq 0. \qquad (*)$$
Thus,
$$\begin{align*} \sum_{i=1}^{n} x_i^n (1 + x_i) &= \sum_{i=1}^{n} x_i^n + \sum_{i=1}^{n} x_i^{n+1} \\ &\geq \sum_{i=1}^{n} x_i^n + n \left( \prod_{i=1}^{n} x_i \right)^{1+1/n} \quad \text{(AM-GM)} \\ &= \sum_{i=1}^{n} x_i^n + n = \sum_{i=1}^{n} (1 + x_i^n) \\ &\geq \frac{1}{2^{n-1}} \sum_{i=1}^{n} (1 + x_i)^n \quad \text{by (*)} \\ &\geq \frac{n}{2^{n-1}} \prod_{i=1}^{n} (1 + x_i). \quad \text{(AM-GM)} \end{align*}$$
Clearly, equality holds if and only if $x_1 = x_2 = \dots = x_n = 1$.