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Algebra Difficulty 4.6 AIME Prove it Ukraine

Let x1,x2,,xnx_1, x_2, \dots, x_n be positive numbers such that:
x1n1+x2n1++xnn1=x1x2xn x_1^{n-1} + x_2^{n-1} + \dots + x_n^{n-1} = x_1 x_2 \dots x_n
Prove the inequality:
(x1n+1)(x2n+1)(xnn+1)1.(Serdiuk Nazar) (x_1 - n + 1)(x_2 - n + 1)\dots(x_n - n + 1) \ge 1. \quad (\text{Serdiuk Nazar})

Solution

Using Cauchy inequality we get i=1,n\forall i=1, n:
x1x2xn=x1n1+x2n1++xnn1x1n1+(n1)x1x2xn1xn+1xnx1x2xn1xn+1xn(xin+1)xin1xin+1xin1x1x2xn1xn+1xni=1n(xin+1)i=1nxin1x1x2xn1xn+1xn=1, \begin{align*} x_1 x_2 \dots x_n &= x_1^{n-1} + x_2^{n-1} + \dots + x_n^{n-1} \ge x_1^{n-1} + (n-1) x_1 x_2 \dots x_{n-1} x_{n+1} \dots x_n \\ &\Rightarrow x_1 x_2 \dots x_{n-1} x_{n+1} \dots x_n (x_i - n + 1) \ge x_i^{n-1} \Rightarrow x_i - n + 1 \ge \frac{x_i^{n-1}}{x_1 x_2 \dots x_{n-1} x_{n+1} \dots x_n} \\ &\Rightarrow \prod_{i=1}^n (x_i - n + 1) \ge \prod_{i=1}^n \frac{x_i^{n-1}}{x_1 x_2 \dots x_{n-1} x_{n+1} \dots x_n} = 1, \end{align*}
as needed.

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