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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.

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  • $\begingroup$ $a,b,c$ are Reals? or precisely natural numbers? $\endgroup$ – Inceptio Mar 15 '13 at 11:37
  • $\begingroup$ Let $x,y,z$ be the natural logarithms of $a,b,c$. Express the inequality in these terms. $\endgroup$ – Macavity Mar 15 '13 at 12:14
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\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.

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  • $\begingroup$ Could you please explain how do you obtain $ \frac{ln(ln(b)) - ln(ln(a))}{ln a} $ from $ \log_a(log_a(b))$ ? Thanks :) $\endgroup$ – GniruT Oct 24 '15 at 11:30
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Let us use the property of $log$.
We know that $\log_a b = \frac{\log_{10} b}{\log_{10}a}$.

If we apply that in your problem, we end up with $$\frac{\log_{10}b}{(\log_{10}a)^2}+\frac{\log_{10}c}{(\log_{10}b)^2}+\frac{\log_{10}a}{(\log_{10}c)^2} $$

Now we can able to see that the quantity is $>0$

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  • $\begingroup$ But $\log_{a}{\log_{a}{b}}=\frac{\log{\log{b}}-\log{\log{a}}}{\log{a}}$ $\endgroup$ – Ivan Loh Mar 15 '13 at 12:23
  • $\begingroup$ @IvanLoh im not sure how to proceed with that identity. $\endgroup$ – Learner Mar 15 '13 at 12:27

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