Question

Prove That
$\sum _ { a , b , c } ( \frac { 1 } { 1 + \log _ { a } b c } ) = 1 \:$
​

Answer 1

Answer:

We know that

$\log_{a}(b) = \frac{ \log(b) }{ \log(a) }$

We have,

$\sum_{a,b,c} ( \frac{1}{1 + \log_{a}(bc) } ) = 1$

Taking LHS

$\frac{1}{1 + \log_{a}(bc) }$

$\frac{1}{1 + \frac{ \log(bc) }{ \log(a) } } = \frac{1}{ \frac{ \log(a) + \log(bc) }{ \log(a) } } = \frac{ \log(a) }{ \log(a) + \log(bc) }$

Using

log( a ) + log( b ) = log (ab)

$= \frac{ \log(a) }{ \log(abc) }$

Thus, the sum is

$= \frac{ \log(a) }{ \log(abc) } + \frac{ \log(b) }{ \log(abc) } + \frac{ \log(c) }{ \log(abc) }$

$= \frac{ \log(a) + \log(b) + \log(c) }{ \log(abc) }$

Using the same formula

$= \frac{ \log(abc) }{ \log(abc) }$

= 1