题目内容
【题目】已知实数x1 , x2 , x3 , x4 , x5满足0<x1<x2<x3<x4<x5
(1)求证不等式x12+x22+x32+x42+x52>x1x2+x2x3+x3x4+x4x5+x5x1
(2)随机变量X取值 的概率均为 ,随机变量Y取值 的概率也均为 ,比较DX与DY大小关系.
【答案】
(1)证明:∵0<x1<x2<x3<x4<x5,
∴ + >2x1x2,
+ >2x2x3,
+ >2x3x4,
+ >2x4x5,
+ >2x5x1,
∴2( + + + + )>2x1x2+2x2x3+2x3x4+2x4x5+2x5x1,
∴x12+x22+x32+x42+x52>x1x2+x2x3+x3x4+x4x5+x5x1
(2)解:设 = (x1+x2+…+x5).
EX= + + + = (x1+x2+…+x5)= .
EY= +…+ = (x1+x2+…+x5)= .
DX= +…+ ,
DY= + +…+ ,
又∵实数x1,x2,x3,x4,x5满足0<x1<x2<x3<x4<x5,
∴DX﹣DY= + +…+
= × ×[(x1﹣x2)(5x1+7x2)+(x2﹣x3)(5x2+7x3)+…+(x5﹣x1)(5x5+7x1)]
= ×(x1x2+x2x3+…+x5x1﹣ ﹣…﹣
=﹣ +…+ <0,
∴DX<DY
【解析】(1)0<x1<x2<x3<x4<x5,利用基本不等式的性质可得 + >2x1x2, + >2x2x3,…, + >2x5x1,相加即可得出.(2)设 = (x1+x2+…+x5).利用数学期望计算公式可得EX,EY.再利用方差计算公式可得DX,DY.作差即可比较出大小关系.