Question

Given a²/(b+c) = b²/(c+a) = c²/(a+b) = 1
Prove that, 1/(1+a) + 1/(1+b) + 1/(1+c) = 1

Answer 1

Question:

If $\sf\frac{a^2}{(b+c)}=\frac{b^2}{(c+a)}=\frac{c^2}{(a+b)}=1$

Prove that,

$\sf\frac{1}{(1+a)}+\frac{1}{(1+b)}+\frac{1}{(1+c)}=1$

Solution:

$\sf\frac{a^2}{(b+c)}=\frac{b^2}{(c+a)}=\frac{c^2}{(a+b)}=1$

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$\sf\frac{a^2}{b+c}=1$

$\implies\sf{a^2=b+c........(i)}$

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$\sf\frac{b^2}{c+a}=1$

$\implies\sf{b^2=c+a.......(ii)}$

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$\sf\frac{c^2}{a+b}=1$

$\implies\sf{c^2=a+b......(iii)}$

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Taking L.H.S,

$\sf\frac{1}{(1+a)}+\frac{1}{(1+b)}+\frac{1}{(1+c)}$

$\implies\sf\frac{a}{a(1+a)}+\frac{b}{b(1+b)}+\frac{c}{c(1+c)}$

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

★Putting all values from (I),(ii)and (iii)no. eq★

$\implies\sf\frac{a}{a+b+c}+\frac{b}{b+c+a}+\frac{c}{c+a+b}$

$\implies\sf\frac{a+b+c}{a+b+c}$

$\implies\sf{1}$

R.H.S = 1

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

Answer 2

$\huge{ \underline{ \underline{ \green{ \sf{ Detailed \: Answer :}}}}}$

•Given :

$\frac{a {}^{2} }{b + c} = \frac{b {}^{2} }{c + a} = \frac{c {}^{2} }{a + b} = 1$

•To prove :

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

Solution :

Given ;

$\frac{a {}^{2} }{b + c} = \frac{b {}^{2} }{c + a} = \frac{c {}^{2} }{a + b} = 1$

On comparing

⇒a²=b+c ...(1)

⇒b²=c+a ....(2)

⇒c²=a+b ....(3)

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LHS

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

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

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

Now put value of equation (1),(2)&(3)

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

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

$= 1$

RHS = 1

⇒LHS = RHS

$\huge{\bold{ Hence\: Proved}}$