若x,y,z都是正实数,且x+y+z=xyz,且1/(x+y)+1/(y+z)+1/(z+x)恒成立,求a的取值范围若x,y,z都_数学解答

若x,y,z都是正实数,且x+y+z=xyz,且1/(x+y)+1/(y+z)+1/(z+x)恒成立,求a的取值范围
若x,y,z都是正实数,且x+y+z=xyz,且1/(x+y)+1/(y+z)+1/(z+x)恒成立,
求a的取值范围
可以用柯西不等式
1/(x+y)+1/(y+z)+1/(z+x)

人气：488 ℃　时间：2020-07-17 08:37:39

解答

[1/(x+y)+1/(y+z)+1/(z+x)]^2≤[1/(x+y)^2+1/(y+z)^2+1/(z+x)^2](1^2+1^2+1^2) （柯西不等式）≤3(1/4xy+1/4yz+1/4zx) (均值不等式)=(3/4)(x+y+z)/xyz=3/4.所以1/(x+y)+1/(y+z)+1/(z+x)≤根号3/2. 且x=y=z=根号3时,...