Question

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

Given :

$\bf{\dfrac{a}{(b + c)} + \dfrac{b}{(c + a)} + \dfrac{c}{(a + b)} = 1}$

To find :

To Proof that ,

$\bf{\dfrac{a^{2}}{(b + c)} + \dfrac{b^{2}}{(c + a)} + \dfrac{c^{2}}{(a + b)} = 0}$

Solution :

$:\implies \bf{\dfrac{a}{(b + c)} + \dfrac{b}{(c + a)} + \dfrac{c}{(a + b)} = 1}$

By multiplying (a + b + c) , we get :

$:\implies \bf{\bigg(\dfrac{a}{(b + c)} + \dfrac{b}{(c + a)} + \dfrac{c}{(a + b)}\bigg) \times (a + b + c) = 1 \times (a + b + c)}$

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

$:\implies \bf{\dfrac{a^{2} + ab + ac}{(b + c)} + \dfrac{ab + b^{2} + cb}{(c + a)} + \dfrac{ac + bc + c^{2}}{(a + b)} = a + b + c}$

We can write , y(x + y + z) as , y² + yx + yz so by using this, we get the Equation as :-

$:\implies \bf{\dfrac{a^{2}}{(b + c)} + \dfrac{ab + ac}{(b + c)} + \dfrac{b^{2}}{(c + a)} + \dfrac{ab + cb}{(c + a)} + \dfrac{c^{2}}{(a + b)} + \dfrac{ac + bc}{(a + b)} = a + b + c}$

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

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

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

$:\implies \bf{\dfrac{a^{2}}{(b + c)} + \dfrac{b^{2}}{(c + a)} + \dfrac{c^{2}}{(a + b)} = a + b + c - (a + b + c)}$

$:\implies \bf{\dfrac{a^{2}}{(b + c)} + \dfrac{b^{2}}{(c + a)} + \dfrac{c^{2}}{(a + b)} = a + b + c - a - b - c}$

$:\implies \bf{\dfrac{a^{2}}{(b + c)} + \dfrac{b^{2}}{(c + a)} + \dfrac{c^{2}}{(a + b)} = \not{a} + \not{b} + \not{c} - \not{a} - \not{b} - \not{c}}$

$:\implies \bf{\dfrac{a^{2}}{(b + c)} + \dfrac{b^{2}}{(c + a)} + \dfrac{c^{2}}{(a + b)} = 0}$

$\boxed{\therefore \bf{\dfrac{a^{2}}{(b + c)} + \dfrac{b^{2}}{(c + a)} + \dfrac{c^{2}}{(a + b)} = 0}}$

Hence,

$\bf{\dfrac{a^{2}}{(b + c)} + \dfrac{b^{2}}{(c + a)} + \dfrac{c^{2}}{(a + b)} = 0}$

Proved !!