I have this statement:
Find the value of $\frac{1}{log_abc+1} + \frac{1}{log_bca+1} + \frac{1}{log_cab+1}, \{a,b,c\} > 0. abc, \{a,b,c\} \neq 1$
My attempt was:
$\frac{1}{log_abc+1} + \frac{1}{log_bca+1} + \frac{1}{log_cab+1} = \sum^{cyc} log_{(bc+1)}a$ for $a,b,c$
If i add $log_{(bc+1)}(bc+1) = 1$ for each element, i.e add three times $1$, i got:
$ \sum^{cyc} log_{(bc+1)}a = [\sum^{cyc}log_{(bc+1)}(abc+a)] - 3$
But i can't get more. Any hint is appreciated.
It seems you are assuming $\log_a{(bc+1)}$ but it seems that it should be $\log_a{(bc)+1}$ and by $\log_A B=\frac{\log A}{\log B}$ we obtain
$$\frac{1}{\log_abc+1} + \frac{1}{\log_bca+1} + \frac{1}{\log_cab+1}=$$
$$\frac{1}{\frac{\log bc}{\log a}+1} + \frac{1}{\frac{\log ca}{\log b}+1} + \frac{1}{\frac{\log ab}{\log c}+1}=$$
$$=\frac{\log a}{\log (bc)+\log a} + \frac{\log b}{\log (ca)+\log b} + \frac{\log c}{\log (ab)+\log c} =$$
$$=\frac{\log a+\log b+\log c}{\log abc}=\frac{\log abc}{\log abc}=1$$
To avoid any possible confusion we should use brackets but when brackets are omitted, by $\log A+B$ we usually mean $\log (A) + B$.
