Suppose that \displaystyle a,b\ \text{and}\ c are real numbers greater than 1 .Find the value of \displaystyle \frac{1}{{1+{{{\log }}_{{{{a}^{2}}b}}}\left( {\frac{c}{a}} \right)}}+\frac{1}{{1+{{{\log }}_{{{{b}^{2}}c}}}\left( {\frac{a}{b}} \right)}}+\frac{1}{{1+{{{\log }}_{{{{c}^{2}}a}}}\left( {\frac{b}{c}} \right)}} .
Solution
\displaystyle \frac{1}{{1+{{{\log }}_{{{{a}^{2}}b}}}\left( {\frac{c}{a}} \right)}}+\frac{1}{{1+{{{\log }}_{{{{b}^{2}}c}}}\left( {\frac{a}{b}} \right)}}+\frac{1}{{1+{{{\log }}_{{{{c}^{2}}a}}}\left( {\frac{b}{c}} \right)}}
\displaystyle =\frac{1}{{{{{\log }}_{{{{a}^{2}}b}}}{{a}^{2}}b+{{{\log }}_{{{{a}^{2}}b}}}\left( {\frac{c}{a}} \right)}}+\frac{1}{{{{{\log }}_{{{{b}^{2}}c}}}{{b}^{2}}c+{{{\log }}_{{{{b}^{2}}c}}}\left( {\frac{a}{b}} \right)}}+\frac{1}{{{{{\log }}_{{{{c}^{2}}a}}}{{c}^{2}}a+{{{\log }}_{{{{c}^{2}}a}}}\left( {\frac{b}{c}} \right)}}
\displaystyle =\frac{1}{{{{{\log }}_{{{{a}^{2}}b}}}abc}}+\frac{1}{{{{{\log }}_{{{{b}^{2}}c}}}abc}}+\frac{1}{{{{{\log }}_{{{{c}^{2}}a}}}abc}}
\displaystyle ={{\log }_{{abc}}}{{a}^{2}}b+{{\log }_{{abc}}}{{b}^{2}}c+{{\log }_{{abc}}}{{c}^{2}}a
\displaystyle ={{\log }_{{abc}}}{{(abc)}^{3}}
\displaystyle =3
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