已知x,y,z∈R^+,x+y+z=xyz,且去1/(x+y)+1/(y+z)+1/(z+x)≤k恒成立,则k的取值范围
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已知x,y,z∈R^+,x+y+z=xyz,且去1/(x+y)+1/(y+z)+1/(z+x)≤k恒成立,则k的取值范围是?
![已知x,y,z∈R^+,x+y+z=xyz,且去1/(x+y)+1/(y+z)+1/(z+x)≤k恒成立,则k的取值范围](/uploads/image/z/8706531-3-1.jpg?t=%E5%B7%B2%E7%9F%A5x%2Cy%2Cz%E2%88%88R%5E%2B%2Cx%2By%2Bz%3Dxyz%2C%E4%B8%94%E5%8E%BB1%2F%28x%2By%29%2B1%2F%28y%2Bz%29%2B1%2F%28z%2Bx%29%E2%89%A4k%E6%81%92%E6%88%90%E7%AB%8B%2C%E5%88%99k%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4)
只须求出1/(x+y)+1/(y+z)+1/(z+x)的最大值即可知道k的范围.
∵x+y+z=xyz
∴1/(xy)+1/(yz)+1/(zx)=1
由柯西不等式知:
[1/(x+y)+1/(y+z)+1/(z+x)]^2
≤(1^2+1^2+1^2)[1/(x+y)^2+1/(y+z)^2+1/(z+x)^2]
而且(x+y)^2=x^2+y^2+2xy≥2xy+2xy=4xy
∴1/(x+y)^2≤1/(4xy)
同理:1/(y+z)^2≤1/(4yz),1/(z+x)^2≤1/(4zx)
∴1/(x+y)^2+1/(y+z)^2+1/(z+x)^2≤[1/(xy)+1/(yz)+1/(zx)]/4=1/4
∴[1/(x+y)+1/(y+z)+1/(z+x)]^2
≤(1^2+1^2+1^2)[1/(x+y)^2+1/(y+z)^2+1/(z+x)^2]
≤3/4
∴1/(x+y)+1/(y+z)+1/(z+x)≤√3/2
以上不等式等号成立的条件是x=y=z=√3
即1/(x+y)+1/(y+z)+1/(z+x)最大值为√3/2
∴当k≥√3/2时,1/(x+y)+1/(y+z)+1/(z+x)≤k恒成立
k∈(√3/2 ,+∞)
∵x+y+z=xyz
∴1/(xy)+1/(yz)+1/(zx)=1
由柯西不等式知:
[1/(x+y)+1/(y+z)+1/(z+x)]^2
≤(1^2+1^2+1^2)[1/(x+y)^2+1/(y+z)^2+1/(z+x)^2]
而且(x+y)^2=x^2+y^2+2xy≥2xy+2xy=4xy
∴1/(x+y)^2≤1/(4xy)
同理:1/(y+z)^2≤1/(4yz),1/(z+x)^2≤1/(4zx)
∴1/(x+y)^2+1/(y+z)^2+1/(z+x)^2≤[1/(xy)+1/(yz)+1/(zx)]/4=1/4
∴[1/(x+y)+1/(y+z)+1/(z+x)]^2
≤(1^2+1^2+1^2)[1/(x+y)^2+1/(y+z)^2+1/(z+x)^2]
≤3/4
∴1/(x+y)+1/(y+z)+1/(z+x)≤√3/2
以上不等式等号成立的条件是x=y=z=√3
即1/(x+y)+1/(y+z)+1/(z+x)最大值为√3/2
∴当k≥√3/2时,1/(x+y)+1/(y+z)+1/(z+x)≤k恒成立
k∈(√3/2 ,+∞)
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