If #a,b,c# are in HP then reciprocal of them should be in AP .
Hence we have the following relation
#1/a+1/c=2/b#
=>#(a+c)/(ac)=2/b......(1)#
=>#b(a+c)=2ac......(2)#
Now we are to show that #a/(b+c),b/(c+a),c/(a+b)# are in HP . If we can show that the reciprocal of these three are in AP, then the it will be proved that the given three quantities are in HP. So we are to prove the following relation
#(b+c)/a+(a+b)/c=(2(c+a))/b#
Now LHS
#=(b+c)/a+(a+b)/c#
#=(bc+c^2+a^2+ab)/(ac)#
#=(b(c+a)+c^2+a^2)/(ac)#
[using relation(2) #color(red)(b(a+c)=2ac)#]
#=(2ac+c^2+a^2)/(ac)#
#=(c+a)^2/(ac)#
#=((c+a)(c+a))/(ac)#
[using relation(1) =>#color(red)((a+c)/(ac)=2/b)#]
#=(2(c+a))/b=RHS#
