比较大小 5x^2+y^2+z^2与2xy+4x+2z-2

5x^2+y^2+z^2－（2xy+4x+2z-2）＝4x^2－4x+1+x^2+y^2-2xy+z^2-2z+1=(2x-1)^2+(x+y)^2+(z-1)^2>=0
所以5x^2+y^2+z^2大于等于2xy+4x+2z-2

( 5x^2 + y^2 + z^2) - (2xy + 4x + 2z - 2)

= (x^2 - 2xy + y^2) + (4x^2 - 4x +1)+(z^2 - 2z +1)

=(x - y)^2 +(2x -1)^2 +(z-1)^2 >= 0

所以有:
当 x = y = 1/2, z=1时,两者相等;其它任何情况都是前者大于后者.