Question

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

Answer 1

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Multiplying both the sides with a + b + c the result will be ( a + b + c is not equal to zero)

$= > \frac{a(a + b + c)}{b + c} + \frac{b(a + b + c)}{c + a} + \frac{c(a + b + c)}{a + b} = (a + b + c) \\ \\ = \frac{ {a}^{2} }{b + c} + \frac{a(b + c)}{b + c} + \frac{ {b}^{2} }{c + a} + \frac{b(c + a)}{c + a} + \frac{ {c}^{2} }{a + b} + \frac{c(a + b)}{a + b} = (a + b + c) \\ \\ = \frac{ {a}^{2} }{b + c} + a + \frac{ {b}^{2} }{c + a} + b + \frac{ {c}^{2} }{a + b} + c = (a + b + c) \\ \\ = \frac{ {a}^{2} }{b + c} + \frac{ {b}^{2} }{c + a} + \frac{ {c}^{2} }{a + b} = 0$

Hence proved,


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Answer 2

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