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Finally an Inequality

Source: 2024 CTST P18

March 25, 2024
2024 CTSTinequalitiesconbinatorics

Problem Statement

Let m,n∈Z≥0,m,n\in\mathbb Z_{\ge 0}, a0,a1,…,am,b0,b1,…,bn∈R≥0a_0,a_1,\ldots ,a_m,b_0,b_1,\ldots ,b_n\in\mathbb R_{\ge 0} For any integer 0≤k≤m+n,0\le k\le m+n, define ck:=max⁡i+j=kaibj.c_k:=\max_{i+j=k}a_ib_j. Proof 1m+n+1∑k=0m+nck≥1(m+1)(n+1)∑i=0mai∑j=0nbj.\frac 1{m+n+1}\sum_{k=0}^{m+n}c_k\ge\frac 1{(m+1)(n+1)}\sum_{i=0}^{m}a_i\sum_{j=0}^{n}b_j.
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