I'm currently working on proving the inequality $$ \frac{\log(\log(m))}{\log(\log(n))} \left(1 + \frac{\log(n/m)}{\log(n) \log(\log(m))}\right) < 1 $$ where $n > m$, and I suspect that using the inequalities $\frac{t}{1+t} < \log(1+t) < t$ could be crucial in the proof.
I've attempted to apply these inequalities to the expression $\frac{\log(\log(m))}{\log(\log(n))}$, but I'm encountering some challenges.
Any insights, suggestions, would be greatly appreciated! Thank you in advance.
Firstly, I think another condition is necessary which is n > e. Here is my proof of the following inequality :
Let us put log(n) = a and log(m) = b ($\frac{b}{a} < 1 \because n > m$) and use $log(\frac {n}{m}) = log(n) - log(m)$
Thus we get $\frac {log(log(m))}{log(log(n))} (1 + \frac {log(\frac {n}{m})}{log(n) log(log(m))}) = \frac {log(b)}{log(a)} (1 + \frac {a - b}{a\log(b)})$
Dividing by a in the numerator we get $\frac {log(b)}{log(a)} (1 + \frac {a - b}{a\log(b)}) = \frac {log(b)}{log(a)} (1 + \frac {1 - \frac {b}{a}}{log(b)})$
Opening the brackets we get $\frac {log(b)}{log(a)} + \frac {(1-\frac{b}{a})log(b)}{log(a)*log(b)} = \frac {log(b)}{log(a)} + \frac {(1-\frac{b}{a})}{log(a)} = \frac {log(b) + 1-\frac{b}{a}}{log(a)}$
To prove that $\frac {log(b) + 1-\frac{b}{a}}{log(a)} < 1$, subtract 1 on both sides
$\frac {log(b) + 1-{b}{a}}{log(a)} - 1 < 0$
$\frac {log(b) + 1-\frac{b}{a} - log(a)}{log(a)} < 0$
(Use log(b) - log (a) = log($\frac {b}{a}$))
$\frac {log(\frac{b}{a}) + 1-\frac{b}{a}}{log(a)} < 0$
if $log(a) > 0 \implies a (= log n) > 1, \implies n > e$
then we need $log(\frac{b}{a}) + 1-\frac{b}{a} < 0$
as you said, we can now use the inequality log(1 + t) < t where t = ($\frac{b}{a}$ - 1) to get that this inequality is true