Prove that $x/((e+x)\ln(e+x)) \leq \ln(\ln(e+x)) \leq x/e$ for all $x > 0$ I've been working on a problem but have been stuck for several hours finishing it.
The problem is to show that
$$
  \frac{x}{(e+x) \ln(e+x)} \leq \ln(\ln(e+x)) \leq \frac{x}{e}
$$
for all $x > 0$.
I proceeded by first integrating each component as the problem suggested, which changed it to proving that
$$
\int_0^x \frac{1}{(e+x) \ln(e+x)} \,\mathrm{d}t
\leq
\int_0^x \frac{1}{(e+t) \ln(e+t)} \,\mathrm{d}t
\leq
\int_0^x \frac{1}{e} \,\mathrm{d}t \,.
$$
From here I was able to show quite easily that when $x>0$, the second inequality holds. However, I'm unsure of how to now prove the first inequality: is there a result which could help with this or something that I'm missing?
Thanks very much for any help, and sorry for the poor formatting, it's my first time using stack exchange.
Mark
 A: Using just the integral definition of $\ln$, we have
$$\ln (\ln(e+x)) = \int_1^{\ln(e+x)} \frac{dt}{t} \geqslant \frac{\ln(e+x) -1}{\ln(e+x)} = \frac{\ln (e \cdot (1+x/e))-1}{\ln(e+x)} = \frac{\ln(1+x/e)}{\ln(e+x)} \\= \frac{1}{\ln(e+x)}\int_1^{1+ x/e} \frac{dt}{t}\geqslant \frac{1}{\ln(e+x)}\frac{x/e}{1+x/e}= \frac{x}{(e+x)\ln(e+x)}$$
A: HINT
Let $u = \ln(e + x)$ (observe that $u\geq 1$). Then one has that
\begin{align*}
\frac{x}{(e + x)\ln(e + x)} \leq \ln(\ln(e + x)) & \Longleftrightarrow \frac{e^{u} - e}{e^{u}u} \leq \ln(u)
\end{align*}
If we let $u = 1$, both the RHS and the LHS equals zero.
Now we can take the difference and differentiate it:
\begin{align*}
f(u) := \frac{1}{u} - \frac{e^{1 - u}}{u} - \ln(u) \Rightarrow f'(u) & = -\frac{1}{u^{2}} + \frac{e^{1-u}u + e^{1-u}}{u^{2}} - \frac{1}{u}\\\\
& = \frac{e^{1 - u}(u  + 1) - (u + 1)}{u^{2}}\\\\
& = \frac{(e^{1 - u} - 1)(u + 1)}{u^{2}}
\end{align*}
Consequently, the proposed relation holds iff $f'(u) < 0$ for $u\geq 1$. Indeed, this is the case (why?).
For the other inequality, consider the same substitution $u = \ln(e + x)$ (once again, $u\geq 1$).
Then we have the following equivalence:
\begin{align*}
\ln(\ln(e + x)) \leq \frac{x}{e} \Longleftrightarrow 
\ln(u) \leq \frac{e^{u} - e}{e}
\end{align*}
If we let $u = 1$, both the RHS and LHS equals zero.
Once again, we are going to take the difference and differentiate it:
\begin{align*}
g(u) := \ln(u) - e^{u - 1} + 1 \Rightarrow g'(u) & = \frac{1}{u} - e^{u-1}
\end{align*}
Analogously, the proposed relation holds iff $g'(u) < 0$ for every $u\geq 1$. Indeed, this is the case (why?).
Hopefully this helps!
A: Let $ f(x) = \ln ( \ln (e+x) ) $.

*

*Show that $ f'(x) = \frac{1}{(e+x)( \log e+x)} $.

*Show that $f(0) = 0 $.

*Show that for $X> x$, $$ \frac{1}{ ( e+X) \log (e + X) } \leq f'(x) \leq \frac{1}{e}.$$

*Consider $ \int_0^X f'(x) \, dx$ and derive the desired inequality
$$ \frac{X}{ ( e+X) \log (e + X) } \leq \ln ( \ln (e+X)) \leq \frac{X}{e}.$$
