HMMT 二月 2010 · 冲刺赛 · 第 27 题
HMMT February 2010 — Guts Round — Problem 27
题目详情
- [ 15 ] Suppose that there are real numbers a, b, c ≥ 1 and that there are positive reals x, y, z such thatx y za + b + c = 4
x y zxa + yb + zc = 6
2 x 2 y 2 zx a + y b + z c = 9 .
What is the maximum possible value of c ?
13 ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 20 FEBRUARY 2010 — GUTS ROUNDth
英文原题
[ 15 ] Suppose that there are real numbers a, b, c ≥ 1 and that there are positive reals x, y, z such that
a x + b y + c z = 4
xa x + yb y + zc z = 6
x 2 a x + y 2 b y + z 2 c z = 9 .
What is the maximum possible value of c ?
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
13 th ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 20 FEBRUARY 2010 — GUTS ROUND
解析
英文解析
- [ 15 ] Suppose that there are real numbers a, b, c ≥ 1 and that there are positive reals x, y, z such thatx y za + b + c = 4
x y zxa + yb + zc = 6
2 x 2 y 2 zx a + y b + z c = 9 .
What is the maximum possible value of c ?
√
Answer: 4 The Cauchy-Schwarz inequality states that given 2 sequences of n real numbers 3
2 2 2 2 2 2 2
x , x , . . . , x and y , y , . . . , y , then ( x + x + . . . + x )( y + y + . . . + y ) ≥ ( x y + x y + . . . + x y )
1 2 n 1 2 n 1 1 2 2 n n
1 2 n 1 2 nx x xx/ 2 y/ 2 z/ 2
1 2 nwith equality holding if and only if = = . . . = . Applying this to { a , b , c } andy y y
1 2 nx/ 2 y/ 2 z/ 2 x y z 2 x 2 y 2 z x y z 2
{ xa , yb , zc } yields ( a + b + c )( x a + y b + z c ) ≥ ( xa + yb + zb ) with equality holdingif and only if x = y = z .
However, equality does hold (both sides evaluate to 36), so x = y = z . The second equation thenx x x 3 / 2 3 / 2 3 / 23
becomes x ( a + b + c ) = 6, which implies x = . Then we have a + b + c = 4. To maximize
√2
3 / 23
c , we minimize a and b by setting a = b = 1. Then c = 2 or c = 4.
13 ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 20 FEBRUARY 2010 — GUTS ROUNDth