a^2+b^2+c^2-ab-bc-ac
=(2a^2+2b^2+2c^2-2ab-2bc-2ac)/2
=[(a^2-2ab+b^2)+(b^2-2bc+c^2)+(c^2-2ac+c^2)]/2
=[(a-b)^2+(b-c)^2+(c-a)^2]/2>=0
所以a^2+b^2+c^2>=ab+bc+ac