做差法x²+y²+z²-(xy+yz+xz)=1/2[2x²+2y²+2z²-(2xy+2yz+2xz)]配方=1/2[(x-y)²+(y-z)²+(z-x)²]≥0所以我们可以做了证明因为1/2[(x-y)²+(y-z)²+(z-x)²]≥...