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

Given,

$\dfrac{1}{1 + a + b^{-1}} + \dfrac{1}{1 + b + c^{-1}} + \dfrac{1}{1 + c + a^{-1}} = 1$

Solving L.H.S :

$\Rightarrow \dfrac{1}{1 + a + b^{-1}} + \dfrac{1}{1 + b + c^{-1}} + \dfrac{1}{1 + c + a^{-1}} \\[1em] \Rightarrow \dfrac{1}{1 + a + \dfrac{1}{b}} + \dfrac{1}{1 + b + \dfrac{1}{c}} + \dfrac{1}{1 + c + \dfrac{1}{a}}$

Multiplying numerator and denominator of first term by b, second term by c, and third term by a we get,

$\Rightarrow \dfrac{1 \times b}{\Big(1 + a + \dfrac{1}{b}\Big) \times b} + \dfrac{1 \times c}{\Big(1 + b + \dfrac{1}{c}\Big) \times c} + \dfrac{1 \times a}{\Big(1 + c + \dfrac{1}{a}\Big) \times a} \\[1em] \Rightarrow \dfrac{b}{(b + ab + 1)} + \dfrac{c}{(c + bc + 1)} + \dfrac{a}{(a + ac + 1)}$

Since abc = 1, c = $\dfrac{1}{ab}$.

Substituting, c = $\dfrac{1}{ab}$, we get :

$\Rightarrow \dfrac{b}{(b + ab + 1)} + \dfrac{\dfrac{1}{ab}}{\Big(\dfrac{1}{ab} + b \Big(\dfrac{1}{ab}\Big) + 1\Big)} + \dfrac{a}{(a + a \Big(\dfrac{1}{ab}\Big) + 1)} \\[1em] \Rightarrow \dfrac{b}{(b + ab + 1)} + \dfrac{\dfrac{1}{ab}}{\Big(\dfrac{1}{ab} + \dfrac{1}{a} + 1\Big)} + \dfrac{a}{\Big(a + \dfrac{1}{b} + 1\Big)} \\[1em] \Rightarrow \dfrac{b}{(b + ab + 1)} + \dfrac{1}{ab \times \Big(\dfrac{1 + b + ab}{ab}\Big)} + \dfrac{a}{\Big(\dfrac{ab + 1 + b}{b}\Big)} \\[1em] \Rightarrow \dfrac{b}{(b + ab + 1)} + \dfrac{1}{\Big({1 + b + ab}\Big)} + \dfrac{ab}{\Big(ab + 1 + b\Big)} \\[1em] \Rightarrow \dfrac{b + 1 + ab}{(b + ab + 1)} \\[1em] \Rightarrow \dfrac{1 + b + ab}{(1 + b + ab)} \\[1em] \Rightarrow 1.$

Hence proved, $\dfrac{1}{1 + a + b^{-1}} + \dfrac{1}{1 + b + c^{-1}} + \dfrac{1}{1 + c + a^{-1}} = 1$.