MathDB

2

Part of 1992 Poland - Second Round

Problems(1)

sum a_i x_i <= sum b_i x_i.

Source: Polish MO Recond Round 1992 p2

9/9/2024
Given a natural number n≥2 n \geq 2 . Let a1,a2,…,an a_1, a_2, \ldots , a_n , b1,b2,…,bn b_1, b_2, \ldots , b_n be real numbers. Prove that the following conditions are equivalent:
- For any real numbers x1≤x2≤…≤xn x_1 \leq x_2 \leq \ldots \leq x_n holds the inequality ∑i=1naixi≤∑i=1nbixi.\sum_{i=1}^n a_i x_i \leq \sum_{i=1}^n b_i x_i.
- For every natural number k∈{1,2,…,n−1} k\in \{1,2,\ldots, n-1\} holds the inequality ∑i=1kai≥∑i=1kbi,   and  ∑i=1nai=∑i=1nbi. \sum_{i=1}^k a_i \geq \sum_{i=1}^k b_i, \ \ \text{ and } \\ \ \sum_{i=1}^n a_i = \sum_{i=1 }^n b_i.
algebrainequalities