My Cyclic Logarithmic Sum

$\displaystyle \sum_{\text{cyclic}(a,b,c)} \left( \dfrac{1}{1+\log_a bc} \right)$

$=\dfrac{1}{1+\log_a bc} + \dfrac{1}{1+\log_b ca} + \dfrac{1}{1+\log_c ab}$

$=\dfrac{1}{1+\log_a bc} + \dfrac{1}{1+\dfrac{\log_a ca}{\log_a b}} + \dfrac{1}{1+\dfrac{\log_a ab}{\log_a c}}$

$=\dfrac{1}{1+\log_a bc} + \dfrac{1}{\left(\dfrac{\log_a b + \log_a c + \log_a a}{\log_a b}\right)} + \dfrac{1}{\left(\dfrac{\log_a c + \log_a a + \log_a b}{\log_a c}\right)}$

$=\dfrac{1}{1+\log_a bc} + \dfrac{\log_a b}{1 + \log_a bc} + \dfrac{\log_a c}{1 + \log_a bc}$

$=\dfrac{1 + \log_a bc}{1 + \log_a bc}$

$=\boxed{1}$

Using $\color{#D61F06}{\log_a b=\dfrac{\ln b}{\ln a}}$ and $\color{#3D99F6}{\ln (ab)=\ln a+\ln b}$ summation can be written as:

$\large{\begin{aligned}&\displaystyle\sum_{\text{cyc}}\dfrac{1}{1+\dfrac{\ln bc}{\ln a}}\\&=\displaystyle\sum_{\text{cyc}}\dfrac{\ln a}{\ln a+\ln bc}\\&=\displaystyle\sum_{\text{cyc}} \dfrac{\ln a}{\ln a+\ln b+\ln c}\\&=\dfrac{\displaystyle\sum_{\text{cyc}}\ln a}{\ln a+\ln b+\ln c}\\&\huge = \boxed 1 \end{aligned}}$

$\displaystyle \sum_{\text{cyclic}(a,b,c)} \frac{1}{1+\log_a (bc)} \\ =\displaystyle \sum_{\text{cyclic}(a,b,c)} \frac{1}{\log_a (a)+\log_a (bc)} \\ \displaystyle = \sum_{\text{cyclic}(a,b,c)} \frac{1}{\log_a (abc)} \\ \displaystyle = \sum_{\text{cyclic}(a,b,c)} \log_{abc} (a) \\ = \log_{abc} (a) + \log_{abc} (b) + \log_{abc} (c) \\ = \log_{abc} (abc) \\ = \boxed{1}$

The equation above is true for any positive values of a,b,c. (guess) So just sub in a=b=c=2 and you get the answer

If it's about solving a problem--then be it $a=b=c$ for this case (take 2 for example-or anything valid for this function).