a+b+c=0
两边平方
a^2+b^2+c^2+2ab+2bc+2ca=0
a^2+b^2+c^2=-(2ab+2bc+2ca)
(a^2+b^2+c^2)^2
=[-(2ab+2bc+2ca)]^2
=4(ab+bc+ca)^2
=4(a^2b^2+b^2c^2+c^2a^2+2ab^2c+2a^2bc+2abc^2)
=4[a^2b^2+b^2c^2+c^2a^2+2abc(a+b+c)]
=4(a^2b^2+b^2c^2+c^2a^2+2abc*0)
=4(a^2b^2+b^2c^2+c^2a^2)
所以a^2b^2+b^2c^2+c^2a^2=(a^2+b^2+c^2)^2/4
所以你的结论有错