已知 a>0,b>0 且 a+b=1,求证:(ax+by)(ay+bx)≥xy

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已知 a>0,b>0 且 a+b=1,求证:(ax+by)(ay+bx)≥xy
已知 a>0,b>0 且 a+b=1,求证:(ax+by)(ay+bx)≥xy

已知 a>0,b>0 且 a+b=1,求证:(ax+by)(ay+bx)≥xy
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