HMMT 二月 2020 · 冲刺赛 · 第 32 题

HMMT February 2020 — Guts Round — Problem 32

[18] Find the smallest real constant α such that for all positive integers n and real numbers 0 = y < 0 y < · · · < y , the following inequality holds: 1 n n n 3 / 2 2 ∑ ∑ ( k + 1) k + 3 k + 3 √ ≥ . α 2 2 y y − y k k k − 1 k =1 k =1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT February 2020, February 15, 2020 — GUTS ROUND Organization Team Team ID#

解析

[18] Find the smallest real constant α such that for all positive integers n and real numbers 0 = y < 0 y < · · · < y , the following inequality holds: 1 n n n 3 / 2 2 ∑ ∑ ( k + 1) k + 3 k + 3 α √ ≥ . 2 2 y y − y k k − 1 k k =1 k =1 Proposed by: Andrew Gu √ 16 2 Answer: 9 Solution: We first prove the following lemma: Lemma. For positive reals a, b, c, d , the inequality 3 / 2 3 / 2 3 / 2 a b ( a + b )

≥ 1 / 2 1 / 2 1 / 2 c d ( c + d ) holds. Proof. Apply H¨ older’s inequality in the form ( ) 2 3 / 2 3 / 2 a b 3
( c + d ) ≥ ( a + b ) . 1 / 2 1 / 2 c d 2 2 2 2 For k ≥ 2, applying the lemma to a = ( k − 1) , b = 8 k + 8 , c = y , d = y − y yields k − 1 k k − 1 3 3 / 2 3 ( k − 1) (8 k + 8) ( k + 3) √
≥ . 2 2 y y k − 1 y − y k k k − 1 We also have the equality 3 / 2 3 (8 · 1 + 8) (1 + 3) √ = . 2 2 y y − y 1 1 0 Summing the inequality from k = 2 to k = n with the equality yields n n n 3 / 2 3 3 3 2 ∑ ∑ ∑ (8 k + 8) ( k + 3) − k n 9( k + 3 k + 3) √ ≥ + ≥ . 2 2 y y y y − y k n k k k − 1 k =1 k =1 k =1 √ 16 2 Hence the inequality holds for α = . In the reverse direction, this is sharp when y = n ( n + 1)( n + n 9 k − 1 2)( n + 3) (so that y = y for k = 2, . . . , n ) and n → ∞ . k − 1 k k +3