Home Competitions USAMO 2020 Problem 6
Let n ≥ 2 n \geq 2 n ≥ 2 be an integer. Let x 1 ≥ x 2 ≥ ⋯ ≥ x n x_{1} \geq x_{2} \geq \cdots \geq x_{n} x 1 ≥ x 2 ≥ ⋯ ≥ x n and y 1 ≥ y 2 ≥ ⋯ ≥ y n y_{1} \geq y_{2} \geq \cdots \geq y_{n} y 1 ≥ y 2 ≥ ⋯ ≥ y n be 2 n 2 n 2 n real numbers such that 0 = x 1 + x 2 + ⋯ + x n = y 1 + y 2 + ⋯ + y n and 1 = x 1 2 + x 2 2 + ⋯ + x n 2 = y 1 2 + y 2 2 + ⋯ + y n 2 \begin{aligned} 0 & =x_{1}+x_{2}+\cdots+x_{n}=y_{1}+y_{2}+\cdots+y_{n} \\ \text { and } 1 & =x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}=y_{1}^{2}+y_{2}^{2}+\cdots+y_{n}^{2} \end{aligned} 0 and 1 = x 1 + x 2 + ⋯ + x n = y 1 + y 2 + ⋯ + y n = x 1 2 + x 2 2 + ⋯ + x n 2 = y 1 2 + y 2 2 + ⋯ + y n 2 Prove that ∑ i = 1 n ( x i y i − x i y n + 1 − i ) ≥ 2 n − 1 \sum_{i=1}^{n}\left(x_{i} y_{i}-x_{i} y_{n+1-i}\right) \geq \frac{2}{\sqrt{n-1}} i = 1 ∑ n ( x i y i − x i y n + 1 − i ) ≥ n − 1 2
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