# Given $1<a<b<c$ prove $\log_a\log_ab+\log_b\log_bc+\log_c\log_ca>0.$

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.

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

\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.

• Could you please explain how do you obtain $\frac{ln(ln(b)) - ln(ln(a))}{ln a}$ from $\log_a(log_a(b))$ ? Thanks :) – GniruT Oct 24 '15 at 11:30

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 $>0$

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