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Algebra Difficulty 7.7 National olympiad, round 2 Prove it Romania

Let nn be a positive integer and let x1,x2,,xnx_1, x_2, \dots, x_n be positive real numbers such that x1x2xn=1x_1x_2 \cdots x_n = 1. Prove that
i=1nxin(1+xi)n2n1i=1n(1+xi). \sum_{i=1}^{n} x_{i}^{n}(1 + x_{i}) \geq \frac{n}{2^{n-1}} \prod_{i=1}^{n}(1 + x_{i}).

Solution

By the power-mean inequality,
1+an(1+a)n2n1,a0.() 1 + a^n \geq \frac{(1 + a)^n}{2^{n-1}}, \quad a \geq 0. \qquad (*)
Thus,
i=1nxin(1+xi)=i=1nxin+i=1nxin+1i=1nxin+n(i=1nxi)1+1/n(AM-GM)=i=1nxin+n=i=1n(1+xin)12n1i=1n(1+xi)nby (*)n2n1i=1n(1+xi).(AM-GM) \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 x1=x2==xn=1x_1 = x_2 = \dots = x_n = 1.

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