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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
Ifab+c+bc+a+ca+b=1thenprovethata2b+c+b2c+a+c2a+b=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 .
ab+c+bc+a+ca+b=1(a+b+c)[ab+c+bc+a+ca+b]=a+b+c…….(i)nowleta2b+c+b2c+a+c2a+b=q(i)givesq+a(b+c)b+c+b(c+a)c+a+c(a+b)a+b=a+b+cq=a2b+c+b2c+a+c2a+b=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
=a2b+c+a+b2c+a+b+c2a+b+c−(a+b+c)a2+ab+acb+c+b2+bc+abc+a+c2+ac+bca+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)(ab+c+bc+a+ca+b−1)=(a+b+c)(1−1)=0

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