将1/(a+1)+1/(b+1)+1/(c+1)=0 两边同乘以(a+1)(b+1)(c+1)得:(b+1)(c+1)+(a+1)(c+1)+(a+1)(b+1)=0ab+bc+ca+2(a+b+c)+3=0又:a+b+c=0,所以:ab+bc+ca=-3(a+1)^2+(b+2)^2+(c+3)^2=a^2+b^2+c^2+2(a+b+c)+3=a^2+b^2+c^2+3=(a+b+c)^2-2(ab+bc+ca)+3=0-2*(-3)+3=9
12