2009/026) If a , b , c are in Arithmetic Progression prove: 1/(b1/2+c1/2),1/(c1/2+a1/2),1/… also in AP.

If a , b , c are in Arithmetic Progression prove : 1 / (b1/2+c1/2) , 1 / (c1/2+a1/2) , 1 / (a1/2+b1/2) in AP?

to get rid of ^1/2

Let a = x^2 , b= y^2 and c = z^2

We need to prove

1/(y+z) , 1/(x+z) and 1/(x+y) are in AP


As x^2 y^2 and z^2 are in AP

So x^2+xy + yz + xz , y^2 + +xy + yz + xz , z^2+xy + yz + xz (adding same to all terms) are in AP

So (x+y)(x+z), (y+x)(y+z), (z+x)(z+y) are in AP by factoring

So 1/(y+z) , 1/(x+z) and 1/(x+y) are in AP by deviding each term by (x+y)(x+z)(y+z)