TITLE: Given $1<a<b<c$ prove $\log_a\log_ab+\log_b\log_bc+\log_c\log_ca>0.$
QUESTION [5 upvotes]: Given $1<a<b<c$ prove $$ \log_a\log_ab+\log_b\log_bc+\log_c\log_ca>0. $$

How to approach problems like this? I tried usual transformations but no help. I guess I have to use characteristic of logarithm function, but I'm not quite sure how to approach it.

REPLY [3 votes]: \begin{align}
& \log_{a}{\log_{a}{b}}+\log_{b}{\log_{b}{c}}+\log_{c}{\log_{c}{a}}>0 \\
& \Leftrightarrow \frac{\ln{\ln{b}}-\ln{\ln{a}}}{\ln{a}}+\frac{\ln{\ln{c}}-\ln{\ln{b}}}{\ln{b}}+\frac{\ln{\ln{a}}-\ln{\ln{c}}}{\ln{c}}>0 \\
& \Leftrightarrow \frac{\ln{\ln{b}}}{\ln{a}}+\frac{\ln{\ln{c}}}{\ln{b}}+\frac{\ln{\ln{a}}}{\ln{c}}>\frac{\ln{\ln{a}}}{\ln{a}}+\frac{\ln{\ln{b}}}{\ln{b}}+\frac{\ln{\ln{c}}}{\ln{c}}
\end{align}
The last inequality is true by rearrangement inequality since $\ln{\ln{c}}>\ln{\ln{b}}>\ln{\ln{a}}$ and $\frac{1}{\ln{a}}>\frac{1}{\ln{b}}>\frac{1}{\ln{c}}$ and the inequality is strict since the numbers are distinct.