已知a,b,x,y∈R+,且ab=4,x+y=1,求证(ax+by)(ay+bx)>=4

问题描述:

已知a,b,x,y∈R+,且ab=4,x+y=1,求证(ax+by)(ay+bx)>=4

(ax+by)(ay+bx)=[(√ax)^2+(√by)^2][(√bx)^2+(√ay)^2]
>=[(√ax*√bx+√by*√ay]^2 (柯西不等式)
=[(√ab)x+(√ab)y]^2 =4(x+y)^2=4
即(ax+by)(ay+bx)>=4 当a=b 时取“=”得证.
希望对你有点帮助!