HMMT 二月 2005 · CALC 赛 · 第 8 题

HMMT February 2005 — CALC Round — Problem 8

If f is a continuous real function such that f ( x − 1) + f ( x + 1) ≥ x + f ( x ) for all x , ∫ 2005 what is the minimum possible value of f ( x ) dx ? 1

解析

If f is a continuous real function such that f ( x − 1) + f ( x + 1) ≥ x + f ( x ) for all x , ∫ 2005 what is the minimum possible value of f ( x ) dx ? 1 Solution: 2010012 Let g ( x ) = f ( x ) − x . Then g ( x − 1) + x − 1 + g ( x + 1) + x + 1 ≥ x + g ( x ) + x, or g ( x − 1) + g ( x + 1) ≥ g ( x ) . But now, g ( x + 3) ≥ g ( x + 2) − g ( x + 1) ≥ − g ( x ) . Therefore ∫ ∫ ∫ a +6 a +3 a +6 g ( x ) dx = g ( x ) dx + g ( x ) dx a a a +3 ∫ a +3 = ( g ( x ) + g ( x + 3)) dx ≥ 0 . a It follows that ∫ ∫ 333 2005 6 n +7 ∑ g ( x ) = g ( x ) dx ≥ 0 , 1 6 n +1 n =0 so that [ ] 2005 ∫ ∫ ∫ 2 2 2005 2005 2005 x 2005 − 1 f ( x ) dx = ( g ( x ) + x ) dx ≥ x dx = = = 2010012 . 2 2 1 1 1 1 Equality holds for f ( x ) = x .