Question

If $abc = 1$, show that $\dfrac{1}{{1 + a + {b^{ - 1}}}} + \dfrac{1}{{1 + b + {c^{ - 1}}}} + \dfrac{1}{{1 + c + {a^{ - 1}}}} = 1$

Answer 1

Hint: To prove that $\dfrac{1}{{1 + a + {b^{ - 1}}}} + \dfrac{1}{{1 + b + {c^{ - 1}}}} + \dfrac{1}{{1 + c + {a^{ - 1}}}} = 1$, we have $abc = 1$. So, using this we need to take $c = \dfrac{1}{{ab}}$ and substitute in the LHS of the given equation. After substituting, take LCM and simplify the equation further and we will get our answer.

Complete step by step solution:
In this question, we are given an equation and we need to prove that it is correct.
The given equation is: $\dfrac{1}{{1 + a + {b^{ - 1}}}} + \dfrac{1}{{1 + b + {c^{ - 1}}}} + \dfrac{1}{{1 + c + {a^{ - 1}}}} = 1$ - - - - - - - - - - - (1)
So, we need to prove that LHS of equation (1) is equal to RHS of the equation (1).
For that, we will take the LHS and simplify it to prove that LHS is equal to RHS.
Also, we are given that $abc = 1$.
Now, let us take the LHS of equation (1). Therefore,
$ \Rightarrow LHS = \dfrac{1}{{1 + a + {b^{ - 1}}}} + \dfrac{1}{{1 + b + {c^{ - 1}}}} + \dfrac{1}{{1 + c + {a^{ - 1}}}}$
Now, inverse means reciprocal. Therefore,
$ \Rightarrow LHS = \dfrac{1}{{1 + a + \dfrac{1}{b}}} + \dfrac{1}{{1 + b + \dfrac{1}{c}}} + \dfrac{1}{{1 + c + \dfrac{1}{a}}}$ - - - - - - - - - - (2)
Now,
$
   \Rightarrow abc = 1 \\
   \Rightarrow c = \dfrac{1}{{ab}} \\
 $
So, substitute this value in equation (2), we get
$
   \Rightarrow LHS = \dfrac{1}{{1 + a + \dfrac{1}{b}}} + \dfrac{1}{{1 + b + \dfrac{1}{{\dfrac{1}{{ab}}}}}} + \dfrac{1}{{1 + \dfrac{1}{{ab}} + \dfrac{1}{a}}} \\
   \Rightarrow LHS = \dfrac{1}{{1 + a + \dfrac{1}{b}}} + \dfrac{1}{{1 + b + ab}} + \dfrac{1}{{1 + \dfrac{1}{{ab}} + \dfrac{1}{a}}} \\
 $
Now, taking LCM in denominator, we get
$
   \Rightarrow LHS = \dfrac{1}{{\dfrac{{b + ab + 1}}{b}}} + \dfrac{1}{{1 + b + ab}} + \dfrac{1}{{\dfrac{{ab + 1 + b}}{{ab}}}} \\
   \Rightarrow LHS = \dfrac{b}{{b + ab + 1}} + \dfrac{1}{{b + ab + 1}} + \dfrac{{ab}}{{b + ab + 1}} \\
 $
Now, all the denominators are the same, so we can add the numerators. Therefore, we get
\[
   \Rightarrow LHS = \dfrac{b}{{b + ab + 1}} + \dfrac{1}{{b + ab + 1}} + \dfrac{{ab}}{{b + ab + 1}} \\
   \Rightarrow LHS = \dfrac{{b + ab + 1}}{{b + ab + 1}} \\
   \Rightarrow LHS = 1 \\
   \Rightarrow LHS = RHS \\
 \]
Hence, we have proved that $\dfrac{1}{{1 + a + {b^{ - 1}}}} + \dfrac{1}{{1 + b + {c^{ - 1}}}} + \dfrac{1}{{1 + c + {a^{ - 1}}}} = 1$.

Note:
Here, we can also prove that LHS is equal to RHS by substituting the value of b equal to 1 divided by $ac$.
$
   \Rightarrow abc = 1 \\
   \Rightarrow b = \dfrac{1}{{ac}} \\
 $
By substituting this value and simplifying further, we will still get LHS equal to RHS.