已知正数x,y,z满足x+y+z=xyz.求不等式1/(x+y) + 1/(y+z) + 1/(z+x)的最大值
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已知正数x,y,z满足x+y+z=xyz.求不等式1/(x+y) + 1/(y+z) + 1/(z+x)的最大值
[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,等号成立.
所以1/(x+y)+1/(y+z)+1/(z+x)的最大值是2分之根号3
≤[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,等号成立.
所以1/(x+y)+1/(y+z)+1/(z+x)的最大值是2分之根号3
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