数列问题，在线等，急

1/a-1/b=1/b-1/c =>b/a-1=1-b/c

(b+c)/a-(c+a)/b=b/a+c/a-c/b-a/b=b/a+c(1/a-1/b)-a/b=b/a+c(1/b-1/c)-a/b=b/a-1+c/b-a/b

(c+a)/b-(a+b)/c=c/b+a(1/b-1/c)-b/c=c/b+a(1/a-1/b)-b/c=c/b-a/b+1-b/c

所以(b+c)/a-(c+a)/b=(c+a)/b-(a+b)/c

由题2/b=1/a+1/c;
2=b/a+b/c;
a*1/a+c*1/c=b/a+b/c;
a(2/b-1/c)+c(2/b-1/a)=b/a+b/c;
组合得
2*a/b+2*c/b=b/a+c/a+a/c+b/c;
即 2*(a+c)/b=(b+c)/a+(a+b)/c;
得证