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Algebra Difficulty 7.0 National olympiad, round 2 Prove it

100. (Original problem, 2006.02.12) Let x1,x2,x3,x4∈R−‾x_{1}, x_{2}, x_{3}, x_{4} \in \overline{\mathbf{R}^{-}}, and x12+x22+x32+x42⩽3x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2} \leqslant 3, then
3x1x2x3x4⩾∑x2x3x4−1⩾∑1⩽i<j⩽4xixj−3⩾3∑x1−9\begin{aligned} 3 x_{1} x_{2} x_{3} x_{4} \geqslant & \sum x_{2} x_{3} x_{4}-1 \geqslant \sum_{1 \leqslant i<j \leqslant 4} x_{i} x_{j}-3 \geqslant \\ & 3 \sum x_{1}-9 \end{aligned}

Equality holds in all the above inequalities if and only if one of x1,x2,x3,x4x_{1}, x_{2}, x_{3}, x_{4} is 0 and the other three are equal to 1.

Solution

100. Proof see "A Chain of Inequalities and Its Proof" by Yang Xuezhi in Mathematics Teaching in Middle Schools (Anhui), 2007, Issue 4.

Conjecture: Let x1,x2,⋯ ,xn∈R−‾x_{1}, x_{2}, \cdots, x_{n} \in \overline{\mathbf{R}^{-}}, and x12+x22+⋯+xn2⩽n−1x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2} \leqslant n-1, prove or disprove:
(1) 12(n−1)(n−2)+(n−1)x1x2⋯xn⩾∑1<i<j<nxixj\frac{1}{2}(n-1)(n-2)+(n-1) x_{1} x_{2} \cdots x_{n} \geqslant \sum_{1<i<j<n} x_{i} x_{j};
(2) n−1+x1x2⋯xn⩾x1+x2+⋯+xnn-1+x_{1} x_{2} \cdots x_{n} \geqslant x_{1}+x_{2}+\cdots+x_{n};
(3) 1+(n−1)x1x2⋯xn⩾∑x2x3⋯xn1+(n-1) x_{1} x_{2} \cdots x_{n} \geqslant \sum x_{2} x_{3} \cdots x_{n}.

Equality holds in all three inequalities if and only if one of x1,x2,⋯ ,xnx_{1}, x_{2}, \cdots, x_{n} is 0 and the other n−1n-1 are all equal to 1. Here ∑x2x3⋯xn\sum x_{2} x_{3} \cdots x_{n} denotes the sum of the products of every n−1n-1 numbers among x1,x2,⋯ ,xnx_{1}, x_{2}, \cdots, x_{n}.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.