已知X,Y,Z都是整数且xy+yz+zx=1,求证x+y+z>=根号3

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已知X,Y,Z都是整数且xy+yz+zx=1,求证x+y+z>=根号3

已知X,Y,Z都是整数且xy+yz+zx=1,求证x+y+z>=根号3
已知X,Y,Z都是整数且xy+yz+zx=1,求证x+y+z>=根号3

已知X,Y,Z都是整数且xy+yz+zx=1,求证x+y+z>=根号3
(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)
=x^2+y^2+z^2+2
=(x^2+y^2)/2+(y^2+z^2)/2+(x^2+z^2)/2+2
≥2[√(x^2*y^2)]/2+2[√(y^2*z^2)]/2+2[√(x^2*z^2)]/2+2
=xy+yz+zx+2
=3
(x+y+z)^2≥3
x+y+z≥√3.

x+y+z》3(xyz)^(1/3)
1=xy+yz+zx》3[(xyz)^2]^(1/3)
两边开方
1》√3*(xyz)^(1/3)
两边乘以√3,得
√3》3(xyz)^(1/3)
于是
x+y+z》√3

二楼的好强....