(3X^2+6)-(2X^2+4X+2)
=3X^2+6-2X^2-4X-2
=X^2-4X+4
=(X-2)^2
当X=2时,(X-2)^2=0
则3X^2+6=2X^2+4X+2
X不等于2时,(X-2)^2>0
则3X^2+6>2X^2+4X+2
所以3X^2+6≥2X^2+4X+2
(3X^2+6)-(2X^2+4X+2)
=3X^2+6-2X^2-4X-2
=X^2-4X+4
=(X-2)^2
当X=2时,(X-2)^2=0
则3X^2+6=2X^2+4X+2
X不等于2时,(X-2)^2>0
则3X^2+6>2X^2+4X+2
所以3X^2+6≥2X^2+4X+2