已知非零实数a、b、c满足a^2+b^2+c^2=1,且a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)

a(1/b+1/c)+ a/a +b(1/c+1/a) + b/b +c(1/a+1/b) + c/c = -3 + 1 + 1 + 1
即
(a+b+c)(1/a+1/b+1/c) = 0
a+b+c = 0 或 1/a+1/b+1/c = 0
通分 (ab+bc+ac)/abc = 0 因为abc 不等于 0 所以ab+bc+ac = 0
(a+b+c)^2 = a^2+b^2+c^2+2ab+2bc+2ac = 1
a+b+c = 1,-1
所以 a+b+c = 0,1,-1