已知a,b>0,且a+b=1,求证:(ax+by)(ay+bx)>=xy
2个回答
展开全部
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
(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
参考资料： baiduzhidao
本回答由提问者推荐
已赞过
已踩过<
评论
收起
你对这个回答的评价是？
展开全部
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
(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
已赞过
已踩过<
评论
收起
你对这个回答的评价是？
推荐律师服务：
若未解决您的问题，请您详细描述您的问题，通过百度律临进行免费专业咨询
