题目

设a≥0,b≥0,a+b=1,且x1、x2为正数，y1=ax1+bx2,y2=bx1+ax2,则y1y2与x1x2的关系是(    )A.y1y2≤x1x2                 B.y1y2≥x1x2               C.y1y2＞x1x2               D.y1y2＜x1x2 答案：解析：y1y2=(ax1+bx2)(bx1+ax2)    =abx12+abx22+x1x2(a2+b2)    ≥2abx1x2+x1x2(a2+b2)    =x1x2(a+b)2=x1x2.    故选B.答案：B

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