Question

a/b+c
$a \b + c + b \c + a + c \a + b = 1 then it prove {?a}^{2}$
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Answer 1

$\large\underline{\sf{Given \:Question - }}$

$\sf \: If \: \dfrac{a}{b + c} + \dfrac{b}{c + a} + \dfrac{c}{a + b} = 1 \: then \: prove \: that \:$

$\sf \: \dfrac{ {a}^{2} }{b + c} + \dfrac{ {b}^{2} }{c + a} + \dfrac{ {c}^{2} }{a + b} = 0$

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:\dfrac{a}{b + c} + \dfrac{b}{c + a} + \dfrac{c}{a + b} = 1$

Multiply both sides by (a + b + c), we get

$\rm \: \dfrac{a(a + b + c)}{b + c} + \dfrac{b(a + b + c)}{c + a} + \dfrac{c(a + b + c)}{a + b} = a + b + c$

$\rm \: \dfrac{ {a}^{2} + a(b + c)}{b + c} + \dfrac{ {b}^{2} + b(a + c)}{a + c} + \dfrac{ {c}^{2} + c(a + b)}{a + b} = a + b + c$

On splitting the terms, we get

$\rm \: \dfrac{ {a}^{2} }{b + c} + \cancel{a} + \dfrac{ {b}^{2} }{c + a} + \cancel{b}+ \dfrac{ {c}^{2} }{a + b} + \cancel{c}= \cancel{a} + \cancel{b} + \cancel{c}$

$\rm \: \dfrac{ {a}^{2} }{b + c} + \dfrac{ {b}^{2} }{c + a} + \dfrac{ {c}^{2} }{a + b} = 0$

${\boxed{\boxed{\bf{Hence, Proved}}}}$