$\log(n) + 1 < \frac{n}{\log(n)}$ for Lagarias inequality In Lagarias paper "An Elementary Problem Equivalent to the Riemann
Hypothesis" he uses the following inequality:
$\log(n) + 1 \leqslant\dfrac{n}{\log(n)}\;\;\;$ for $\;n\geqslant3\;.$
I've tried hard, but I couldn't find a proof for this statement. I know is true but I couldn't use roots bounds for this case. How could this inequality be proven?
 A: Let $x \in [3,\infty)$. Then, by substituting $x=e^u$ with $u \in [\log(3),\infty)$, we have
\begin{align}
\log(x)+1 \leq \frac{x}{\log(x)} &\iff \log(x)^2+\log(x) \leq x\\
&\iff \log(e^u)^2+\log(e^u) \leq e^u\\
&\iff u^2+u \leq e^u.
\end{align}
Consider $x\geq4$, i.e. $u \geq \log(4)$ and define $f(u) := e^u$ and $g(u) := u^2+u$. Then, you can check that $f(\log(4)) = 4 \geq \log(4)^2+\log(4) = g(\log(4))$. Moreover, we have $f'(u) = e^u$ and $g'(u) = 2u+1$ and you can check that $f'(u) \geq g'(u)$ for all $u \geq \log(4)$ It follows that the inequality $u^2+u\leq e^u$ holds for all $u \geq \log(4)$, i.e. the original inequality holds for all $x \geq 4$, so in particular for all $n \in \mathbb{N}_{\geq4}$.
You can check manually, that the inequality holds for the case $n=3$ as well. Then you are done. Hope this helps!
A: This is just a minor variant on jasnee's answer. It's not hard to show that $1-{1\over2}u^2+{1\over6}u^3$ is positive for all $u\ge0$. (The derivative is $-u+{1\over2}u^2={1\over2}u(u-2)$, so there's a local minimum at $u=2$, and the cubic is positive there.)  It follows that
$$u+u^2\le1+u+{1\over2}u^2+{1\over6}u^3\le e^u$$
for $u\ge0$ and thus, letting $u=\log x$, we have $\log x+(\log x)^2\le x$ for $x\ge1$.
A: Actually we can prove that
$\log x+1<\dfrac x{\log x}\quad$ for all $\;x\in\mathbb R\;\land\;x>1\;.$
For any $\;x\in\mathbb R\;\land\;x>1\;,\;$ by letting $\;y=\log x>0\;,\;$ it results that
$\begin{align}
\log x+1&=y+1\leqslant y+1+\dfrac{(y+1)(y-2)^2}{6y}=\\
&=\dfrac{6y^2+6y+\big(y+1\big)\left(y^2-4y+4\right)}{6y}=\\
&=\dfrac{6y^2+6y+y^3-4y^2+4y+y^2-4y+4}{6y}=\\
&=\dfrac{4+6y+3y^2+y^3}{6y}<\dfrac{6+6y+3y^2+y^3}{6y}=\\
&=\dfrac1y\left(1+y+\dfrac{y^2}2+\dfrac{y^3}6\right)<\dfrac{e^y}y=\dfrac x{\log x}\;.
\end{align}$
