HMMT 二月 2014 · 代数 · 第 7 题

英文解析

7. Find the largest real number c such that
   ∑101
   2 2
   x ≥ c mii =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.
   Note that f ( t ) = t is a convex function over the reals, so we may “smooth” to the case x = · · · =251
   x 0 ≤ x = · · · = x (the x = · · · is why we needed to assume x ≥ 0). Indeed, by Jensen’s 1
   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 = · · · =
   x = − 51 r and x = · · · = x = 50 r for some r ≥ 0, and by homogeneity, C works if and only if 1
   50 51 101
   2 2
   50(51) +51(50) 51(101)
   5151
   C ≤ = = .
   50 50 502
   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 ) =
   ∑ii
   ( x − x ) . Indeed, we may set u = x − ( x + · · · + x ) / 50 and v = ( x + · · · + x ) / 50 − x ,2
   j i 51 1 50 52 101 51
   2 2 2
   and use the fact that ( u − v ) ≤ u + v .
   4 3 2