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if a2/(b+c)=b2/(c+a)=c2/(a+b)=1, then 1/(1+a)+1/(1+b)+1/(1+c)=1

if a2/(b+c)=b2/(c+a)=c2/(a+b)=1, then 1/(1+a)+1/(1+b)+1/(1+c)=1

Grade:12th pass

2 Answers

Meet
137 Points
6 years ago
As if u solve the equation in 3 variable then u will get that a=2,b=2,c=2. and now if u put these values in 2nd equation then it will become 1/3+1/3+1/3=1 hence proved
mycroft holmes
272 Points
6 years ago
 a^2=b+c \Rightarrow a^2+a = a(1+a) = a+b+c \Rightarrow \dfrac{1}{1+a} = \dfrac{a}{a+b+c}
 
Similarly \dfrac{1}{1+b} = \dfrac{b}{a+b+c} and \dfrac{1}{1+c} = \dfrac{c}{a+b+c}
 
Hence \dfrac{1}{1+a}+ \dfrac{1}{1+b}+ \dfrac{1}{1+c} = \dfrac{a}{a+b+c}+ \dfrac{b}{a+b+c}+\dfrac{c}{a+b+c}
 
\dfrac{a+b+c}{a+b+c}=1
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