Question

if a/b+c +b/c+a +c/a+b =1 prove A2/b+c +b2/c+a +C2/a+b=0​

Answer 1

Step-by-step explanation:

$\frac{a}{b + c} + \frac{b}{c + a} + \frac{c}{a + b} = 1$

therefore,

a = b + c

b = c + a

c = a + b

Now,

$\frac{ {a}^{2} }{b + c} + \frac{ {b}^{2} }{c + a} + \frac{ {c}^{2} }{a + b}$

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

$= ( \frac{ {a}^{2} + ab + ac }{b + c} ) + ( \frac{ {b}^{2} + bc + ab }{c + a} ) + ( \frac{ {c}^{2} + ac + bc }{a + b} ) - (a + b + c)$

$= \frac{a(a + b + c)}{b + c} + \frac{b(b + c + a)}{c + a} + \frac{c(c + a + b)}{a + b} - (a + b + c)$

$= (a + b + c)( \frac{a}{b + c} + \frac{b}{c + a} + \frac{c}{a + b} ) - (a + b + c)$

$= (a + b + c)( \frac{a}{a} + \frac{b}{b} + \frac{c}{c} ) - (a + b + c)$

$= (a + b + c) \times 1 - (a + b + c)$

$= 0$

L. H. S = R. H. S (proved)