Problem 6 Day 2Answer Let n≥2n \geq 2n≥2 be an integer. Let x1≥x2≥⋯≥xnx_{1} \geq x_{2} \geq \cdots \geq x_{n}x1≥x2≥⋯≥xn and y1≥y2≥⋯≥yny_{1} \geq y_{2} \geq \cdots \geq y_{n}y1≥y2≥⋯≥yn be 2n2 n2n real numbers such that 0=x1+x2+⋯+xn=y1+y2+⋯+yn and 1=x12+x22+⋯+xn2=y12+y22+⋯+yn2\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=x1+x2+⋯+xn=y1+y2+⋯+yn=x12+x22+⋯+xn2=y12+y22+⋯+yn2 Prove that ∑i=1n(xiyi−xiyn+1−i)≥2n−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(xiyi−xiyn+1−i)≥n−12 Answer, solutions and source