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if a2/(b+c) +b2/(c+a) +c2/(a+b)=0 then prove that a/(b+c) +b/(c+a) +c/(a+b) =1

if a2/(b+c) +b2/(c+a) +c2/(a+b)=0 then prove that a/(b+c) +b/(c+a) +c/(a+b) =1

Grade:12th pass

2 Answers

Arun
25750 Points
6 years ago
Dear student
 
take a/(b+c) +b/(c+a) +c/(a+b) =1
 
then multiply (a+b+c)  in both the sides
a(a+b+c) / (b+c) + b(a+b+c) / (a+c) + c(a+b+c) / (b+a) = a+b+c
 
a2 / (b+c)  + a(b+c)/ (b+c)  + b2 / (c+a) + b(c+a)/ (c+a)  +  c2 / (a+b)  + c(a+b)/ (a+b) = a+b+c
 
a2 / (b+c)  + a + b2 / (c+a) + b  +  c2 / (a+b)  + c = a+b+c
 
a2 / (b+c)  + b2 / (c+a) +  c2 / (a+b)  = 0
 
Hence Proved
 
 
Regards
Arun (askIITians forum expert)
Sayantan Garai
117 Points
5 years ago
a²/b+c + b²/c+a + c²/a+b = 0
=> a²/b+c + a + b²/c+a + b + c²/a+b + c = a+b+c
=> a²+ab+ac/b+c + b²+bc+ab/c+a + c²+ac+bc/a+b = a+b+c
=> (a+b+c)(a/b+c + b/c+a + c/a+b ) =   a+b+c
=> a/b+c + b/c+a + c/a+b = 1

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