已知x+y+z=a,求证x^2+y^2+z^2>=a^2/3
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已知x+y+z=a,求证x^2+y^2+z^2=a^2/3证:(x+y+z)^2=a^2,即x^2+y^2+z^2+2xy+2yz+2zx=a^2 x^2+y^2+z^2=a^2-(2xy+2yz+2zx),∴x^2+y^2+z^2≥a^2-[(x^2+y^2)+(y^2+z^2)+(z^2+x^2)],即x^2+y^2+z^2≥a^2-2(x^2+y^2+z^2),即3(x^2+y^2+z^2)≥a^2∴x^2+y^2+z^2≥a^2/3