已知正数a,b,c成等差数列,且公差d不等于0,求证:1/a,1/c.1/b不可能成等差数列.

a,b,c成等差数,不妨设b=a+d,c=a+2d则1/a=1/a,1/b=1/(a+d),1/c=1/(a+2d)假设1/a,1/b,1/c能构成等差数列则2/b=1/a+1/c即2/(a+d)=1/a+1/(a+2d)2/(a+d)=(2a+2d)/(a(a+2d))2a(a+2d)=(a+d)(2a+2d)2a(a+2d)=2(a+d)²a(a...