If a2+2bc,b2+2ac,c2+2ab are in A.P. ,show that 1/(b-c),1/(c-a),1/(a-b) are also in A.P.

  1/ b-c , 1/ c-a , 1/ a-b are in AP iff 

 2/ c-a = 1/a-b + 1/ b-c

            = b-c + a-b / (a-b)(b-c

  2/ c-a = (a-c )/ (a-b )(b-c) 

2(a-b)(b-c)= -(c-a)2 

2ab - 2ac - 2b2 + 2bc = -c2 - a2 + 2ac

  2(b2 + 2ac )=  (c2 + 2ab +  a2 + 2bc  )

 which means 

        which is true

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