Question

if abc= 1, then the value of $( 1 + a + \frac{1}{b})^{-1} + ( 1 + a + \frac{1}{c})^{-1} + ( 1 + a + \frac{1}{a})^{-1}$ is?

Answer 1

given,
abc = 1

1/( 1 + a + 1/b) + 1/( 1 + b + 1/c) + 1/(1 +c + 1/a)
=?

abc = 1

let
a =K
b=1
c =1/K
then,
1/( 1 + a + 1/b) = 1/( 1 + K + 1) = 1/( K+2)

1/( 1 + b + 1/c ) = 1/( 1 + 1 + 1/1/K) = 1/(K+2)

1/(1 + c + 1/a) = 1/( 1 + 1/K + 1/K) =K/(K+2)

now add all of these ,

1/( 1 + a + 1/b) + 1/( 1 + b + 1/c) + 1/( 1 + c + 1/a) = 1/(K +2) + 1/(K+2) + K/(K +2)

= ( 1 + 1 + K)/(K + 2)

= ( K + 2)/(K + 2)

=1 ( answer)