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)=-3,求a+b+c的值

因为a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)=-3 所以a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)+3=0 a(1/a+1/b+1/c)+b(1/a+1/b+1/c)+c(1/a+1/b+1/c)=0 (a+b+c)(ab+bc+ca)/abc=0 若a+