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Question Number 35080 by math1967 last updated on 15/May/18
If (a/(b+c)) +(b/(c+a)) +(c/(a+b))=1 then prove that (a^2 /(b+c)) +(b^2 /(c+a)) +(c^2 /(a+b))=0
$${If}\:\frac{{a}}{{b}+{c}}\:+\frac{{b}}{{c}+{a}}\:+\frac{{c}}{{a}+{b}}=\mathrm{1}\:{then}\:{prove}\:{that} \\ $$$$\frac{{a}^{\mathrm{2}} }{{b}+{c}}\:+\frac{{b}^{\mathrm{2}} }{{c}+{a}}\:+\frac{{c}^{\mathrm{2}} }{{a}+{b}}=\mathrm{0} \\ $$
Answered by ajfour last updated on 15/May/18
(a/(b+c))+(b/(c+a))+(c/(a+b))=1 (a+b+c)[(a/(b+c))+(b/(c+a))+(c/(a+b))]=a+b+c .......(i) now let (a^2 /(b+c))+(b^2 /(c+a))+(c^2 /(a+b))= q (i) gives q+((a(b+c))/(b+c))+((b(c+a))/(c+a))+((c(a+b))/(a+b))=a+b+c q= (a^2 /(b+c))+(b^2 /(c+a))+(c^2 /(a+b)) = 0 .
$$\frac{{a}}{{b}+{c}}+\frac{{b}}{{c}+{a}}+\frac{{c}}{{a}+{b}}=\mathrm{1} \\ $$$$\left({a}+{b}+{c}\right)\left[\frac{{a}}{{b}+{c}}+\frac{{b}}{{c}+{a}}+\frac{{c}}{{a}+{b}}\right]={a}+{b}+{c} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…….\left({i}\right) \\ $$$${now}\:{let}\:\:\frac{{a}^{\mathrm{2}} }{{b}+{c}}+\frac{{b}^{\mathrm{2}} }{{c}+{a}}+\frac{{c}^{\mathrm{2}} }{{a}+{b}}=\:{q} \\ $$$$\left({i}\right)\:{gives} \\ $$$${q}+\frac{{a}\left({b}+{c}\right)}{{b}+{c}}+\frac{{b}\left({c}+{a}\right)}{{c}+{a}}+\frac{{c}\left({a}+{b}\right)}{{a}+{b}}={a}+{b}+{c} \\ $$$$\:\:\:{q}=\:\frac{{a}^{\mathrm{2}} }{{b}+{c}}+\frac{{b}^{\mathrm{2}} }{{c}+{a}}+\frac{{c}^{\mathrm{2}} }{{a}+{b}}\:=\:\mathrm{0}\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 15/May/18
=(a^2 /(b+c))+a+(b^2 /(c+a))+b+(c^2 /(a+b))+c−(a+b+c) ((a^2 +ab+ac)/(b+c))+((b^2 +bc+ab)/(c+a))+((c^2 +ac+bc)/(a+b))−(a+b+c) =((a(a+b+c))/(b+c))+((b(a+b+c))/(c+a))+((c(a+b+c))/(a+b))−(a+b+c) =(a+b+c)((a/(b+c))+(b/(c+a))+(c/(a+b))−1) =(a+b+c)(1−1) =0
$$=\frac{{a}^{\mathrm{2}} }{{b}+{c}}+{a}+\frac{{b}^{\mathrm{2}} }{{c}+{a}}+{b}+\frac{{c}^{\mathrm{2}} }{{a}+{b}}+{c}−\left({a}+{b}+{c}\right) \\ $$$$\frac{{a}^{\mathrm{2}} +{ab}+{ac}}{{b}+{c}}+\frac{{b}^{\mathrm{2}} +{bc}+{ab}}{{c}+{a}}+\frac{{c}^{\mathrm{2}} +{ac}+{bc}}{{a}+{b}}−\left({a}+{b}+{c}\right) \\ $$$$=\frac{{a}\left({a}+{b}+{c}\right)}{{b}+{c}}+\frac{{b}\left({a}+{b}+{c}\right)}{{c}+{a}}+\frac{{c}\left({a}+{b}+{c}\right)}{{a}+{b}}−\left({a}+{b}+{c}\right) \\ $$$$=\left({a}+{b}+{c}\right)\left(\frac{{a}}{{b}+{c}}+\frac{{b}}{{c}+{a}}+\frac{{c}}{{a}+{b}}−\mathrm{1}\right) \\ $$$$=\left({a}+{b}+{c}\right)\left(\mathrm{1}−\mathrm{1}\right) \\ $$$$=\mathrm{0} \\ $$

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