Question Number 40236 by scientist last updated on 17/Jul/18
$${if}\:{a}\:{b}\:{c}\:{are}\:{in}\:{H}.{P}\:{then}\:{prove}\:{that}\:\frac{{a}}{{b}+{c}},\frac{{b}}{{c}+{a}},\frac{{c}}{{a}+{b}}\:\:{arw} \\ $$$${also}\:{in}\:{H}.{P}. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 17/Jul/18
$${a},{b},{c}\:\:{H}.{P} \\ $$$${so}\:\frac{\mathrm{1}}{{a}},\frac{\mathrm{1}}{{b}},\frac{\mathrm{1}}{{c}}\:{in}\:{A}.{P} \\ $$$$\frac{\mathrm{2}}{{b}}=\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{c}} \\ $$$${to}\:{prove}\:\frac{{a}}{{b}+{c}},\frac{{b}}{{c}+{a}},\frac{{c}}{{a}+{b}}\:{in}\:{H}.{P} \\ $$$${if}\:{we}\:{can}\:{prove}\:{reciprocal}\:{are}\:{in}\:{A}.{P} \\ $$$$\frac{{b}+{c}}{{a}},\frac{{c}+{a}}{{b}},\frac{{a}+{b}}{{c}}\:\:{are}\:{in}\:{A}.{P} \\ $$$${now}\:\frac{\mathrm{1}}{{a}},\frac{\mathrm{1}}{{b}},\frac{\mathrm{1}}{{c}}\:\:{are}\:{in}\:{A}.{P} \\ $$$${multiply}\:{a}+{b}+{c}\:{with}\:{each}\:{term}..{results}\:{are}\: \\ $$$${in}\:{A}.{P} \\ $$$$\frac{{a}+{b}+{c}}{{a}},\frac{{a}+{b}+{c}}{{b}},\frac{{a}+{b}+{c}}{{c}}\:\:{in}\:{A}.{P} \\ $$$$\mathrm{1}+\frac{{b}+{c}}{{a}},\mathrm{1}+\frac{{a}+{c}}{{b}},\mathrm{1}+\frac{{a}+{b}}{{c}}\:\:{inA}.{P} \\ $$$$\frac{{b}+{c}}{{a}},\frac{{a}+{c}}{{b}},\frac{{a}+{b}}{{c}}\:\:{inA}.{P} \\ $$$$\frac{{a}}{{b}+{c}},\frac{{b}}{{a}+{c}},\frac{{c}}{{a}+{b}}\:\:{are}\:{in}\:{H}.{P}\:\:\:\:{proved} \\ $$$$ \\ $$