查人人中国名人网广场舞李琦简历:数学题目
来源：百度文库 编辑：查人人中国名人网 时间：2024/05/01 16:39:15
已知x>0.y>0,求证
(1/2)*(x+y)^2+(1/4)*(x+y)>=x*根号y+y*根号x
(1/2)*(x+y)^2+(1/4)*(x+y)>=x*根号y+y*根号x
有没有学基本不等式啊
左边=(1/2)x^2+xy+(1/2)y^2+(1/4)x+(1/4)y
=x^2+(1/4)y+y^2+(1/4)x-(1/2)x^2+xy-(1/2)y^2
>=2根号[x^2+(1/4)y]+2根号[y^2+(1/4)x]-(1/2)*(x-y)^2
=x*根号y+y*根号x-(1/2)*(x-y)^2
又因为(1/2)*(x-y)^2>=0
所以左边>=x*根号y+y*根号x
即(1/2)*(x+y)^2+(1/4)*(x+y)>=x*根号y+y*根号x
A