Question

a^2/b+c=b^2/c+a=c^2/a+b=1 then show that (1/1+a)+(1/1+b)+(1/1+c)=1

Answer 1

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

From the above equalities,

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

Similarly,

$b=\frac{a+b+c}{1+b}.....................(2)\\ c=\frac{a+b+c}{1+c}......................(3)$

Now add equations, (1)+(2)+(3)

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

The proof is complete.