已知x,y,z>0,xyz(x+y+z)=1,求证(x+y)(x+z)>=2
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已知x,y,z>0,xyz(x+y+z)=1,求证(x+y)(x+z)>=2
![已知x,y,z>0,xyz(x+y+z)=1,求证(x+y)(x+z)>=2](/uploads/image/z/18989870-14-0.jpg?t=%E5%B7%B2%E7%9F%A5x%2Cy%2Cz%3E0%2Cxyz%28x%2By%2Bz%29%3D1%2C%E6%B1%82%E8%AF%81%EF%BC%88x%2By%EF%BC%89%28x%2Bz%29%3E%3D2)
证明:由x,y,z>0及均值不等式可知,(yz)+[1/(yz)]≥2.将xyz(x+y+z)=1代替该不等式左端的1.可得(yz)+x(x+y+z)≥2.===>x²+(y+z)x+yz≥2.===>(x+y)(x+z)≥2.
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