已经a+b+c=0求证a(b分之1+c分之1）+b（c分之1+a分之1）+c（a分之1+b分之1）+3=0

左边=a(1/b+1/c)+a/a+b(1/c+1/a)+b/b+c(1/a+1/b)+c/c=a(1/a+1/b+1/c)+b(1/a+1/b+1/c)+c(1/a+1/b+1/c)=(a+b+c)(1/a+1/b+1/c)=0=右边

1/b+1/c=(b+c)/(bc)=(-a)/(bc)
1/c+1/a=(a+c)/(ac)=(-b)/(ac)
1/a+1/b=(a+b)/(ab)=(-c)/(ab)
左边=(-a)/(bc)+（-b)/(ac)+(-c)/(ab)+3=-(a²+b²+c²)/(abc)+3
=-[(a+b+c)²-2ab-2ac-2bc]/(abc)+3=(2ab+2ac+2bc)/(abc)+3
=0

把3分成b/b+a/a+c/c你很快会发现答案；或者a+b=-c换元（都换）也很简单

a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)+3
=(a/b+a/c)+(b/c+b/a)+(c/a+c/b)+a/a+b/b+c/c
=a/a+b/a+c/a+a/b+b/b+c/b+a/c+b/c+c/c
=(a+b+c)/a+(a+b+c)/b+(a+b+c)/c
=(a+b+c)(1/a+1/b+1/c)=0