If b+a/a,c+a/b, a+b/c are in AP show that 1/a,1/b,1/c are in AP

if $\frac{b+c}{a} , \frac{c+a}{b} ,\frac{a+b}{c}$  are  in AP
then $\frac{b+c}{a} +1,\frac{c+a}{b} +1,\frac{a+b}{c} +1$ are  also  in  AP
$\Longrightarrow \frac{b+c+a}{a} ,\ \frac{c+a+b}{b} ,\frac{a+b+c\ }{c}$   are  also  in AP
then   $\frac{c+a+b}{b} -\frac{b+c+a}{a} =\frac{a+b+c\ }{c} -\frac{c+a+b}{b}$  [ condition  for  AP]
$\frac{1}{b} -\frac{1}{a} =\frac{1}{c} -\frac{1}{b}$
which implies that $\frac{1}{a} ,\frac{1}{b}$   and  $\frac{1}{c}$  are  in AP