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PUMaC 2014 · 个人决赛（A 组） · 第 2 题

PUMaC 2014 — Individual Finals (Division A) — Problem 2

专题
Discrete Math / 离散数学
难度
L3
来源
PUMaC

题目详情

  1. Given a, b, c ∈ R , and that a + b + c = 3. Prove that 1 1 1
    • ≥ 1 3 3 3 a + 2 b + 2 c + 2
解析
  1. Given a, b, c 2 R , and that a + b + c = 3. Prove that 1 1 1
    • 1 3 3 3 a + 2 b + 2 c + 2 Solution: 1 ✓ ◆ 3 1 1 a 3 We see that = 1 . By AM GM on the denominator, we have a +1+1 3 3 a + 2 2 a + 2 ✓ ◆ 3 2 1 a 1 a 3 a . Hence 1 . 3 2 a + 2 2 3 2 2 2 1 1 1 3 a + b + c Therefore, we have + + = 1 3 3 3 a + 2 b + 2 c + 2 2 3