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,

[tex] \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) [/tex]

Similarly,

[tex] b=\frac{a+b+c}{1+b}.....................(2)\\

c=\frac{a+b+c}{1+c}......................(3) [/tex]

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

[tex] 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 [/tex]

The proof is complete.