TITLE: Something about $\frac{\log x}{x}$
QUESTION [6 upvotes]: Denote $\log x = \log_ex$. Let's consider the below function $$\frac{\log x}{x}$$. Apparently, It's maximum is $\frac{1}{e}$. and strictly increasing in $(0,e]$, strictly decreasing in $[e,+\infty)$. If we draw a line $y=a$, where $0<a<\frac{1}{e}$. It will have two intersection point $x_1,x_2$. the question is how to prove $$x_1x_2 > e^2$$

REPLY [2 votes]: Consider $x_1$ and $x_2$ as functions of $y\in(0,\frac1e)$, defined implicitly by $y=\frac{\log x_1}{x_1}$ and $y=\frac{\log x_2}{x_2}$ and $x_1<e<x_2$.  Differentiating $\log x = xy$ implicitly wrt $y$ yields $\frac{x'}{x} = x + x'y$, and so
$$ x' = \frac{x}{\frac1x-y} $$
Thus
\begin{align*}
(x_1x_2)'
&= x_1' x_2 + x_1 x_2' \\
&= \frac{x_1x_2}{\frac1{x_1}-y} + \frac{x_1x_2}{\frac1{x_2}-y} \\
&= \frac{x_1x_2\big(\frac1{x_1}+\frac1{x_2} - 2y\big)}{\big(\frac1{x_1}-y\big)\big(\frac1{x_2}-y\big)} 
\end{align*}
Since $x_1<e<x_2$, we have
$$\frac1{x_1} - y = \frac1{x_1}(1-\log x_1) > 0 $$
and
$$\frac1{x_2} - y = \frac1{x_2}(1-\log x_2) < 0 $$
and, by the inequalities of the harmonic, geometric, and logarithmic means,
\begin{align*}
\frac1{x_1}+\frac1{x_2} - 2y
&> \frac2{\sqrt{x_1x_2}} - 2y \\
&> \frac{2(\log x_1 - \log x_2)}{x_1-x_2} - 2y \\
&= \frac{2(x_1y - x_2y)}{x_1-x_2} - 2y \\
&= 0
\end{align*}
(The inequalities are strict because $x_1\ne x_2$.)  Putting that all together yields $(x_1x_2)' < 0$, so $x_1x_2$ is a strictly decreasing function of $y\in (0,\frac1e)$; thus the value at $y=\frac1e$ is the (unique global) minimum, that is, $x_1x_2 > e^2$.