Question

已知b^2=ac,求证:1/(a+b) ,1/2b,1/(b+c)成等差数列

Answer 1

要证 1/(a+b) ,1/2b,1/(b+c)成等差数列

只须证 1/(b+c)-1/2b=1/2b-1/(a+b)

只须证 (2b-b-c)/(2b(b+c))=(a+b-2b)/(2b(a+b))

即 (b-c)/(b+c)=(a-b)/(a+b)

即 b^2=ac

由已知 b^=ac

则得证