HMMT 二月 2017 · 冲刺赛 · 第 32 题

HMMT February 2017 — Guts Round — Problem 32

[ 17 ] Let a, b, c be non-negative real numbers such that ab + bc + ca = 3. Suppose that 9 3 3 3 a b + b c + c a + 2 abc ( a + b + c ) = . 2 3 3 3 What is the minimum possible value of ab + bc + ca ? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . February 2017, February 18, 2017 — GUTS ROUND Organization Team Team ID#

解析

[ 17 ] Let a, b, c be non-negative real numbers such that ab + bc + ca = 3. Suppose that 9 3 3 3 a b + b c + c a + 2 abc ( a + b + c ) = . 2 3 3 3 What is the minimum possible value of ab + bc + ca ? Proposed by: Pakawut Jiradilok Answer: 18 ∑ 2 Expanding the inequality ab ( b + c − 2 a ) ≥ 0 gives cyc ( ∑ ) ( ∑ ) ( ∑ ) 3 3 2 2 ab + 4 a b − 4 a b − abc ( a + b + c ) ≥ 0 cyc cyc cyc (∑ ) 9 3 Using a b + 2 abc ( a + b + c ) = in the inequality above yields 2 cyc ( ∑ ) ( ∑ ) ( ∑ ) 3 2 3 2 2 ab − 4( ab + bc + ca ) ≥ ab − 4 a b − 9 abc ( a + b + c ) ≥ − 18 cyc cyc cyc ∑ 3 Since ab + bc + ca = 3, we have ab ≥ 18 as desired. cyc √ √ 3 The equality occurs when ( a, b, c ) ∼ ( , 6 , 0). 2 cyc