三式相加
(a+b+c)x^2+(a+b+c)x+(a+b+c)=0
(a+b+c)(x^2+x+1)=0
x^2+x+1>0
a+b+c=0
a2/bc+b2/ca+c2/ab
=(a^3+b^3+c^3)/abc
=[a(b+c)^2+b(a+c)^2+c(a+b)^2]/abc
=(ab^2+ac^2+ba^2+bc^2+ca^2+cb^2+6abc)/abc
=(ab^2+ac^2+ba^2+bc^2+ca^2+cb^2)/abc+6
=b/c+c/b+a/c+c/a+a/b+b/a +6
=1/b(a+c)+1/c(a+b)+1/a(b+c) +6
=-b/b-c/c-a/a +6
=3
ax^2+bx+c=0 bx^2+cx+a=0 cx^2+ax+b=0
ax^2+bx+c+ bx^2+cx+a+ cx^2+ax+b=0
(a+b+c)[x^2+x+1]=0
x^2+x+1>0
a+b+c=0
(a^3+b^3+c^3-3abc)=(a+b+c)[aa+bb+cc-ab-ac-bc]
a^2/bc + b^2/ac +c^2/ab =(a^3+b^3+c^3-3abc)/abc+3=3