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

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Solution

$\frac{b}{b + a b + 1} + \frac{c}{c + c b + 1} + \frac{a}{a + a c + 1}$

$c = \frac{1}{a b} o r \frac{1}{c} = a b$

$= \frac{b}{b + a b + 1} + \frac{1}{1 + b + a b} + \frac{a b}{1 + a b + b}$

$= \frac{b + 1 + a b}{1 + a b + b} = \begin{matrix} 1 = 1 L . H . S = R . H . S \end{matrix}$

Hence proved.

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

If ABC=1 , show that 1/1+a+b^-1 + 1/1+b+c^-1 + 1/1+c+a^-1=1

a b c = 1

\frac{1}{1 + a + b^{- 1}} + \frac{1}{1 + b + c^{- 1}} + \frac{1}{1 + c + a^{- 1}} = 1

\frac{a + b + c}{a^{- 1} b^{- 1} + b^{- 1} c^{- 1} + c^{- 1} a^{- 1}} = a b c

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