因x,y,z属于R
所以
x^2+y^2≥2xy
x^2+z^2≥2xz
y^2+z^2≥2yz
相加得
2(x^2+y^2+z^2)≥2(xy+yz+zx)
所以x^2+y^2+z^2≥xy+yz+zx
∵(x-y)²≥0,(y-z)²≥0,(z-x)²≥0
∴x²+y²≥2xy,y²+z²≥2yz,z²+x²≥2zx
∴(x²+y²)+(y²+z²)+(z²+x²)≥2xy+2yz+2zx
∴2(x²+y²+z²)≥2(xy+yz+zx)
∴(x²+y²+z²)≥(xy+yz+zx)
x²+y²≥2xy;y²+z²≥2yz;z²+x²≥2xz,三个式子相加,得:
2x²+2y²+2z²≥2xy+2yz+2zx
即:
x²+y²+z²≥xy+yz+zx
(x-y)^2+(y-z)^2+(z-x)^2≥0
x^2-2xy+y^2+y^2-2yz+z^2+z^2-2zx+x^2≥0
2(x^2+y^2+z^2)-2(xy+yz+zx)≥0
2(x^2+y^2+z^2)≥2(xy+yz+zx)
x^2+y^2+z^2≥xy+yz+zx