Show $\log (\log n + 1) \leq \log \log n + \frac{1}{\log (n + 1)}$ How does one show that $\log (\log n + 1) \leq \log \log n + \dfrac{1}{\log (n + 1)}$ for $n \geq 3$?
For context, I saw this inequality in https://arxiv.org/pdf/math/0008177v2 (formula 3.8).
 A: Let's show that

For $x\in\left[0,\frac{1}{2}\right]$ we have
  $$\ln(1+x) \leq x-\frac{x^2}{3} \tag{1}$$

Indeed, $f(x)=\ln(1+x)-x + \frac{x^2}{3}$ has first derivative $f'(x)=\frac{2x}{3}+\frac{1}{1+x}-1=\frac{x(2x-1)}{3(x+1)}\leq 0$ when $x\in \left[0,\frac{1}{2}\right]$.
So $f(x)$ is descending on $x\in \left[0,\frac{1}{2}\right]$ or for $0\leq x\leq \frac{1}{2} \Rightarrow f(0)\geq f(x)$ which is $0\geq \ln(1+x)-x + \frac{x^2}{3}$ and the result follows.
This can also be reformulated as, for $x \geq 2$
$$\ln\left(1+\frac{1}{x}\right) \leq \frac{1}{x}-\frac{1}{3x^2} \iff 
\color{blue}{x\ln\left(1+\frac{1}{x}\right) \leq 1-\frac{1}{3x}} \tag{2}$$

I will also use the fact that
$$\left(1+\frac{1}{x}\right)^x<e \tag{3}$$

Now let's change the original inequality as
$$\log (\log n + 1) \leq \log \log n + \frac{1}{\log (n + 1)}\iff \\
\log\left(1+\frac{1}{\log{n}}\right)\leq \frac{1}{\log(n+1)} \iff \\
\log(n+1)\cdot\log\left(1+\frac{1}{\log{n}}\right)\leq 1 \iff \\
\left(\log{n}+\log\left(1+\frac{1}{n}\right)\right)\cdot\log\left(1+\frac{1}{\log{n}}\right)\leq 1 \iff $$
$$\left(\log{n}+\frac{\log\left(1+\frac{1}{n}\right)^n}{n}\right)\cdot\log\left(1+\frac{1}{\log{n}}\right)\leq 1 \tag{4}$$

And finally
$$\left(\log{n}+\frac{\log\left(1+\frac{1}{n}\right)^n}{n}\right)\cdot\log\left(1+\frac{1}{\log{n}}\right) \overset{(3)}{<} \\
\left(\log{n}+\frac{1}{n}\right)\cdot\log\left(1+\frac{1}{\log{n}}\right) = \\
\left(1+\frac{1}{n\log{n}}\right)\cdot \color{red}{\log{n} \cdot\log\left(1+\frac{1}{\log{n}}\right)} \overset{(2)}{<} \\
\left(1+\frac{1}{n\log{n}}\right)\cdot \left(1-\frac{1}{3\log{n}}\right)=\\
1-\frac{1}{3n\log^2(n)}-\frac{1}{3\log{n}}+\frac{1}{n\log{n}} < 1$$
since 
$$-\frac{1}{3\log{n}}+\frac{1}{n\log{n}} <0$$
and for $\log{n}>2 \iff n >e^2 > 7$. So $(4)$ is true for $n>7$. Cases $n\in\{3,4,5,6,7\}$ can be validated manually or with a computer program.
A: One way is to prove that
$\min\left(\log\left(1+\frac{1}{\log n}\right)\right) \leq \max \left(\frac{1}{\log n+1}\right)$, 
which can easily be done by differentiating both expressions and equating them to $0$. 
But in this case, both derivatives don't come out to be $0$ at any point. So just finding their tendency value as $n$ tends to the extreme values of the domain, should give you their respective extrema.
