已知a,b>0,且a+b=1,求证:(ax+by)(ay+bx)>=xy
问题描述：
已知a,b>0,且a+b=1,求证:(ax+by)(ay+bx)>=xy
答
a+b=1
(a+b)²=1
a²+2ab+b²=1
a²+b²=1-2ab
(ax+by)(ay+bx)
=a²xy+abx²+aby²+b²xy
=ab(x²+y²)+(a²+b²)xy
=ab(x²+y²)+(1-2ab)xy
=ab(x²+y²)-2abxy+xy
=ab(x²-2xy+y²)+xy
=ab(x-y)²+xy
∵a、b∈R+,即ab＞0,且(x-y)²≥0
∴ab(x-y)²≥0,即ab(x-y)²+xy≥xy
∴(ax+by)(ay+bx)≥xy