3(a^2+b^2+c^2)-(a+b+c)^2 =3(a^2+b^2+c^2)-a^2+b^2+c^2+2ab+2bc+2ac =2(a^2+b^2+c^2)-(2ab+2bc+2ac) =(a-b)^2+(b-c)^2+(a-c)^2=0 平方都大于等于0相加为0则各项均为0所以a=b=c
