Question

if abc = 1,prove that:
1/(1+a+b^-1) + 1/(1+b+c^-1) + 1/(1+c+a^-1) =1

Answer 1

Answer:

hope it helps u..........

Answer 2

$\displaystyle \sf{ \frac{1}{1 + a + {b}^{ - 1} } +\frac{1}{1 + b + {c}^{ - 1} } + \frac{1}{1 + c + {a}^{ - 1} } } = 1 \: \: is \: \: \bf{proved}$

Given :

abc = 1

To find :

To prove

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

Solution :

Step 1 of 2 :

Find the value of c

abc = 1

$\displaystyle \sf{ \implies c = \frac{1}{ab} }$

Step 2 of 2 :

Prove the expression

$\displaystyle \sf{ \frac{1}{1 + a + {b}^{ - 1} } +\frac{1}{1 + b + {c}^{ - 1} } + \frac{1}{1 + c + {a}^{ - 1} } }$

$\displaystyle \sf{ = \frac{1}{1 + a + \frac{1}{b} } +\frac{1}{1 + b + \frac{1}{c} } + \frac{1}{1 + c + \frac{1}{a} } }$

$\displaystyle \sf{ = \frac{1}{ \frac{b + ab + 1}{b} } +\frac{1}{1 + b + ab } + \frac{1}{1 + \frac{1}{ab} + \frac{1}{a} } }$

$\displaystyle \sf{ = \frac{b}{1 + b + ab } +\frac{1}{1 + b + ab } + \frac{1}{ \frac{ab + 1 + b}{ab} } }$

$\displaystyle \sf{ = \frac{b}{1 + b + ab } +\frac{1}{1 + b + ab } + \frac{ab}{1 + b + ab } }$

$\displaystyle \sf{ = \frac{b + 1 + ab}{1 + b + ab } }$

$\displaystyle \sf{ = \frac{1 + b + ab}{1 + b + ab } }$

$\displaystyle \sf{ = 1}$

Hence the proof follows

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