How do I go about proving that for all $x\geq 3$, $\log \log (x) \leq \log (\log(x-1)) + 1$?
When I differentiate to see if the lhs stays ahead, i lose the constant on the lhs and so i dont get anything meaningful. I also tried using some known inequalities like Jensen for concave functions but a naive application gives out an inequality in the other direction which is quite useless for this problem.
Any help is appreciated, thanks!
Best Answer
Assume in this answer that $\log$ means the natural logarithm with base $e$.
Since $\log A-\log B = \log\frac{A}{B}$, your inequality is equivalent to $$ \log \frac{\log(x-1)}{\log (x)}=\log (\log(x-1))-\log \log (x)\ge -1=\log\frac{1}{e}\;, $$ which is, by monotonicity of $\log$: $$ \frac{\log(x-1)}{\log (x)}\ge \frac{1}{e}\;. $$ So you want to show that for all $x\ge 3$: $$ f(x) = e\log(x-1)-\log(x)\geq 0\;. $$ Now, for all $x\ge 3$: $$ f'(x) = \frac{e}{x-1}-\frac{1}{x} = \frac{(e-1)x+1}{x(x-1)}\;>0 $$ But $$ f(3) = e\log 2 - \log 3>0. $$