(5x²+y²+z²)-(2xy+4x+2z-2)=(x²-2xy+y²)+(4x²-4x+1)+(z²-2z+1)=(x-y)²+(2x-1)²+(z-1)²因x,y,z为实数，故(x-y)²、(2x-1)²、(z-1)²均不小于0即：(x-y)²+(2x-1)²+(z-1)²≥0所以：(5x²+y²+z²)-(2xy+4x+2z-2)≥0即：5x²+y²+z²≥2xy+4x+2z-2
