Touching Logarithm After Long Days

I like to swap bases instead of using a new base : $\log_a{b}=\frac 1{\log_b{a}}$

$\begin{aligned} S & = \frac 1{1+\log_{a^2b} (\frac ca)} + \frac 1{1+\log_{c^2a} (\frac bc)} + \frac 1{1+\log_{b^2c} (\frac ab)} \\ &= \frac 1{\log_{a^2b}(a^2b) + \log_{a^2b} (\frac ca)} + \frac 1{\log_{c^2a}(c^2a)+\log_{c^2a} (\frac bc)} + \frac 1{\log_{b^2c}(b^2c)+\log_{b^2c} (\frac ab)} \\ &= \frac 1{\log_{a^2b}(abc)} + \frac 1{\log_{c^2a}(abc)} + \frac 1{\log_{b^2c}(abc)} \\ &= \log_{abc}(a^2b) + log_{abc}(b^2c)+\log_{abc}(c^2a) \\ &= \log_{abc}(abc)^3 \\ &= \boxed{3} \end{aligned}$

$\begin{aligned} S & = \frac 1{1+\log_{a^2b} \frac ca} + \frac 1{1+\log_{c^2a} \frac bc} + \frac 1{1+\log_{b^2c} \frac ab} \\ & = \frac 1{1+\frac {\log c - \log a}{2\log a + \log b}} + \frac 1{1+\frac {\log b - \log c}{2\log c + \log a}} + \frac 1{1+\frac {\log a - \log b}{2\log b + \log c}} \\ & = \frac {2\log a + \log b}{2\log a + \log b+ \log c - \log a} + \frac {2\log c + \log a}{2\log c + \log a+ \log b - \log c} + \frac {2\log b + \log c}{2\log b + \log c + \log a - \log b} \\ & = \frac {2\log a + \log b}{\log a + \log b+ \log c} + \frac {2\log c + \log a}{\log a + \log b+ \log c} + \frac {2\log b + \log c}{\log a + \log b+ \log c} \\ & = \frac {3\log a + 3\log b+ 3\log c}{\log a + \log b+ \log c} \\ & = \boxed{3} \end{aligned}$