已知非零实数abc满足a2+b2+c2=1,a(1/b+1/c)+b(1/a+1/c)+c(1/a+1/b)=-3求a+b+c

Question

已知非零实数abc满足a2+b2+c2=1,a(1/b+1/c)+b(1/a+1/c)+c(1/a+1/b)= -3求a+b+c

Answer 1

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+b+c=0,则问题得解.
若ab+bc+ca=0,又因为(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)
故(a+b+c)^2=1+0=1
a+b+c=1或-1 .