若x1/y1＝x2/y2＝……＝xn/yn,求证：根号x1y1＋根号x2＋y2＋……＋根号xnyn＝根号下（x1＋x2＋

Marketing_ff2022-10-04 11:39:541条回答

若x1/y1＝x2/y2＝……＝xn/yn,求证：根号x1y1＋根号x2＋y2＋……＋根号xnyn＝根号下（x1＋x2＋……＋xn)(y1＋y2＋……＋yn)

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老狼是我共回答了11个问题 | 采纳率81.8%: 设x1,x2,……,xn>=0;y1,y2,……,yn>0,
求证：根号x1y1＋根号x2y2＋……＋根号xnyn＝根号下（x1＋x2＋……＋xn)(y1＋y2＋……＋yn)
证：设xi/yi=a>=0,i=1,2,……,n
左=(x1y1)^(1/2)+(x2/y2)^(1/2)+……+(xn/yn)^(1/2)
=[a(y1)^2]^(1/2)+[a(y2)^2]^(1/2)+……+[a(yn)^2]^(1/2)
=a^(1/2)(y1+y2+……+yn)]
右=[（x1＋x2＋……＋xn)(y1＋y2＋……＋yn)]^(1/2)
=[a(y1＋y2＋……＋yn)(y1＋y2＋……＋yn)]^(1/2)=a^(1/2)(y1+y2+……+yn)]
左=右证毕; 1年前