Answer:
Step-by-step explanation:
abc = 1
We have to prove that,
[tex]\frac{1}{1+a+b^{-1}}+\frac{1}{1+b+c^{-1}}+\frac{1}{1+c+a^{-1}}=1[/tex]
We take left hand side of the given equation and solve it,
[tex]\frac{1}{1+a+\frac{1}{b}}+\frac{1}{1+b+\frac{1}{c}}+\frac{1}{1+c+\frac{1}{a}}[/tex]
Since, abc = 1,
[tex]\frac{1}{c}=ab[/tex] and c = [tex]\frac{1}{ab}[/tex]
By substituting these values in the expression,
[tex]\frac{1}{1+a+\frac{1}{b}}+\frac{1}{1+b+\frac{1}{c}}+\frac{1}{1+c+\frac{1}{a}}=\frac{1}{1+a+\frac{1}{b}}+\frac{1}{1+b+ab}+\frac{1}{1+\frac{1}{ab}+\frac{1}{a}}[/tex]
[tex]=\frac{b}{b+ab+1}+\frac{1}{1+b+ab}+\frac{ab}{ab+1+b}[/tex]
[tex]=\frac{1+b+ab}{1+b+ab}[/tex]
[tex]=1[/tex]
Which equal to the right hand side of the equation.
Hence, [tex]\frac{1}{1+a+b^{-1}}+\frac{1}{1+b+c^{-1}}+\frac{1}{1+c+a^{-1}}=1[/tex]
