Question

1/（a+b）,1/(a+c),1/(b+c),是等差数列,求证,a^2,b^2,c^2成等差数列

Answer 1

1/(a+b)+1/(b+c)=2/(a+c) ==>(b+c)(a+c)+(a+b)(a+c)=2(a+b)(b+c)

所以 ba+bc+ac+c^2+a^2+ac+ab+bc=2ab+2ac+2b^2+2bc

整理得：a^2+c^2=2b^2 ,即 a^2,b^2,c^2成等差数列