Question

设X Y Z属于R,求证x²+y²+z²≥xy+yz+zx

Answer 1

∵2(x²+y²+z²)-2(xy+yz+xz)

=(x²+y²-2xy)+(y²+z²-2yz)+(x²+z²-2xz)

=(x-y)²+(y-z)²+(x-z)²

≥0

∴2(x²+y²+z²)≥2(xy+yz+xz)

∴x²+y²+z²≥xy+yz+zx