已知X,Y,Z都是整数且xy+yz+zx=1,求证x+y+z>=根号3
人气：144 ℃ 时间：2020-04-02 04:05:16
解答
（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≥3x+y+z≥√3.
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