题目

(1)已知a,b是正数,且a+b=1.求证:(ax+by)(ay+bx)≥xy.(2)若x+3y-1=0,求证:2x+8y≥. 答案：证明:(1)左边=(a2+b2)xy+ab(x2+y2)=［(a+b)2-2ab］xy+ab(x2+y2)=(1-2ab)xy+ab(x2+y2)=xy+ab(x-y)2,∵a＞0,b＞0,(x-y)2≥0,∴左边≥xy=右边.因此不等式成立.(2)2x+23y≥.

查看本题答案和解析