(2)若x+3y-1=0,求证:2x+8y≥
.
(2)若x+3y-1=0,求证:2x+8y≥.
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≥.