Is it true that for $x>0$, one has $x\le\frac{1}{\log x}$?

Question

I know that for $x>0$, one can quite easily prove that

$$\frac{1}{x}\le\frac{1}{\log{x}},$$

which follows from the trivial identity that $x\le e^x\iff \log x\le x\iff\frac{1}{\log x}\ge\frac{1}{x}$,

but what about the inequality in the title:

$$x\le\frac{1}{\log x}$$

Answer 1

You can try to find the domain on which your inequality holds. For sure we have to require $x>1$, since for $x\leq 1$ we have $\log x\leq 0$.

Then since $x$ is increasing and $1/\log x$ is decreasing for $x>1$, the equality will holds true for $x\in (1,x_\star)$, where $x_\star$ is the only point such that $$x_\star = \frac{1}{\log x_\star}\,.$$

The exact value od $x_\star$ can be found numerically and it's $x_\star = 1.76322\dots$

You can try to find analytic lower bounds for $x_\star$. For instance you know that for $x>1$, $\frac{1}{x-1}\leq \frac{1}{\log x}$ so that $x_\star<x_0$, where $x_0 = \frac{1}{x_0-1}$, i.e. $x_0 = \frac{1}{2}(1+\sqrt{5})=1.61803\dots$. From here you clearly have a rougher bound $$x_\star > x_0=\frac{1}{2}(1+\sqrt{5})>\frac{1}{2}(1+2)=\frac{3}{2}=1.5\,.$$

On the other hand, you have that $x>1+\log x$, so an upper bound for $x_\star$ is the point $x_1$ such that $1+\log x_1=\frac{1}{\log x_1}$. You can find $x_1 = e^{\frac{1}{2}(\sqrt{5}-1)}=1.85528\dots$ As before a less tight but simpler bound can be derived. $$x_\star < x_1=e^{\frac{1}{2}(\sqrt{5}-1)}<e\,.$$

Answer 2

$x\le e^x\iff \log x\le x\not \iff\frac{1}{\log x}\ge\frac{1}{x}$ (NOT true for negative values of $\log x$).

If $\log x < 0 < x$ then $\frac 1{\log x} < \frac 1x$. So $\frac 1{\log x} \le \frac 1{\log x}$ is not true if $x < 1$. But it is true from $x \ge 1$.

And for $x \ge 1$, in order to have $\log x \le x \le \frac 1{\log x}$ we must have $\log x \le \frac 1{\log x}$ which for non-negative $\log x\ge 0$ would mean $\log x \le 1$ and $x \le e$. So for $1\le x \le e$ we have $\frac \log x \le x \le \frac 1{\log x}$ but for $x > e$ we have $\frac 1x \le \frac 1{\log x} < 1 < \log x < x$.

As for $x < 1$ when we have $\log x < 0$ we have $\frac 1{\log x} < 0 < \frac 1x$. If $\frac 1e \le x \le 1$ then we have $\frac 1{\log x} \le -1 \le \log x < 0<\frac 1e \le x < 1 < e \le \frac 1x$.

If $0 < x < \frac 1e$ we have $\log x < -1 < \frac 1{\log x} < 0 < x < \frac 1e < 1 < e < \frac 1x$.