若实数x,y,z满足x^2+y^2+z^2=1,则xy+yz+zx的取值范围是?[-1/2,1]
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![若实数x,y,z满足x^2+y^2+z^2=1,则xy+yz+zx的取值范围是?[-1/2,1]](/uploads/image/z/1343997-45-7.jpg?t=%E8%8B%A5%E5%AE%9E%E6%95%B0x%2Cy%2Cz%E6%BB%A1%E8%B6%B3x%5E2%2By%5E2%2Bz%5E2%3D1%2C%E5%88%99xy%2Byz%2Bzx%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4%E6%98%AF%3F%5B-1%2F2%2C1%5D)
若实数x,y,z满足x^2+y^2+z^2=1,则xy+yz+zx的取值范围是?[-1/2,1]
若实数x,y,z满足x^2+y^2+z^2=1,则xy+yz+zx的取值范围是?
[-1/2,1]
若实数x,y,z满足x^2+y^2+z^2=1,则xy+yz+zx的取值范围是?[-1/2,1]
x^2+y^2+z^2-ab-ac-bc=1/2[(a-c)^2+(b-c)^2+(a-b)^2]
>=0
则1-(ab+bc+ac)>=0
ab+bc+ac=0
则 1+2(ab+bc+ac)>=0
ab+bc+ac
设xy+yz+zx=t
2[x^2+y^2+z^2-(xy+yz+zx)]=(x-y)^2+(x-z)^2+(y-z)^2
2(1-t)=(x-y)^2+(x-z)^2+(y-z)^2>=0
t<=1
2[x^2+y^2+z^2+xy+yz+zx]=(x+y)^2+(x+z)^2+(y+z)^2
2(1+t)=(x+y)^2+(x+z)^2+(y+z)^2>=0
t>=-1/2
所以t:[-1/2,1]