HMMT 二月 2014 · 代数 · 第 7 题
HMMT February 2014 — Algebra — Problem 7
题目详情
- Find the largest real number c such that 101 X 2 2 x cM i i =1 whenever x , . . . , x are real numbers such that x + · · · + x = 0 and M is the median of x , . . . , x . 1 101 1 101 1 101 4 3 2
解析
- Find the largest real number c such that 101 ∑ 2 2 x ≥ cM i i =1 whenever x , . . . , x are real numbers such that x + · · · + x = 0 and M is the median of 1 101 1 101 x , . . . , x . 1 101 5151 1 Answer: OR 103 . 02 OR 103 Suppose without loss of generality that x ≤ · · · ≤ x and 1 101 50 50 M = x ≥ 0. 51 2 Note that f ( t ) = t is a convex function over the reals, so we may “smooth” to the case x = · · · = 1 x 0 ≤ x = · · · = x (the x = · · · is why we needed to assume x ≥ 0). Indeed, by Jensen’s 5 51 101 51 51 x + ··· + x x + ··· + x 1 50 1 50 inequality, the map x , x , . . . , x → , . . . , will decrease or fix the LHS, while 1 2 50 50 50 preserving the ordering condition and the zero-sum condition. Similarly, we may without loss of generality replace x , . . . , x with their average (which will decrease 51 101 or fix the LHS, but also either fix or increase the RHS). But this simplified problem has x = · · · = 1 x = − 51 r and x = · · · = x = 50 r for some r ≥ 0, and by homogeneity, C works if and only if 50 51 101 2 2 50(51) +51(50) 51(101) 5151 C ≤ = = . 2 50 50 50 Comment: One may also use the Cauchy-Schwarz inequality or the QM-AM inequality instead of Jensen’s inequality. ∑ ∑ 2 2 Comment: For this particular problem, there is another solution using the identity 101 x − ( x ) = i i ∑ 2 ( x − x ) . Indeed, we may set u = x − ( x + · · · + x ) / 50 and v = ( x + · · · + x ) / 50 − x , j i 51 1 50 52 101 51 2 2 2 and use the fact that ( u − v ) ≤ u + v . 4 3 2