Something about $\frac{\log x}{x}$ 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 $$x_1x_2 > e^2$$
 A: I would change the independent variable to $t=\log x$. Then we consider the function $f(t)=te^{-t}$, which has maximum at $t=1$. (Indeed, $f'(t)=(1-t)e^{-t}$.) We take intersection points $t_1,t_2$ with horizontal line $y=a>0$, and want to prove that $t_1+t_2>2$. 
Let's say $t_1<1<t_2$. Geometrically, $t_1+t_2>2$   means  that of these two points, $t_2$ is further away from $1$. Which is intuitively clear, since $|f'|$, the rate of change of $f$, is smaller to the right of $1$. So it takes longer for $f$ to decrease  from $1/e$ to $a$ when going to the right. 
A formal computation to justify the above can look like this: 
$$
e^{-1}-a = \int_{t_1}^1 (1-t) e^{-t}\,dt  >  e^{-1} \int_{t_1}^1 (1-t) \,dt
=  \frac{(1-t_1)^2}{2e}
$$
$$
e^{-1}-a = \int_{1}^{t_2} (t-1) e^{-t}\,dt  < e^{-1} \int_{1}^{t_2}  (t-1) \,dt =  \frac{(t_2 - 1)^2}{2e}
$$
A: Turn this around.
Suppose $1 < x < e$
and
$y = e^2/x$.
We want to show that
$\ln y/y > \ln x/x$.
If this holds,
since
$\ln x/x$ is decreasing
for $x > e$,
then the value of $z> e$ such that
$\ln z/z = \ln x/x$
must satisfy
$z > y$
so that
$z > e^2/x$
or $xz > e^2$.
$\ln y/y
=\ln(e^2/x)/(e^2/x)
=x(2-\ln x)/e^2
$
so we want
$x(2-\ln x)/e^2
> \ln x/x$
or
$x^2(2-\ln x)
> e^2 \ln x
$.
Let
$f(x)
=x^2(2-\ln x)
- e^2 \ln x
$
where
$1 < x < e$.
We want to show that
$f(x) > 0$.
I'm going to look 
at $f$ and its successive derivatives
and,
I hope,
show what I want.
Here goes.
$f(1) = 2$
and
$f(e)
=e^2-e^2
=0
$,
so the extreme values are OK.
$\begin{array}\\
f'(x)
&=2x(2-\ln x)-x-e^2/x\\
&=4x-2x\ln x - x -e^2/x\\
&=3x-2x\ln x -e^2/x\\
\end{array}
$
If we can show $f'(x) < 0$ 
we are done.
$f'(1) = 3-e^2 < 0$
and
$f'(e)
=3e-2e-e
=
0
$
so,
again,
the extreme values are ok.
Let $g(x) = f'(x)= 3x-2x\ln x -e^2/x$.
If we can show that
$g(x)$ is increasing,
then we are done.
$\begin{array}\\
g'(x)
&=3-(2+2\ln x)+e^2/x^2\\
&=1-2\ln x+e^2/x^2\\
\end{array}
$
$g'(1) = 1+e^2 > 0$
and
$g'(e) = 0$
so the extreme values are OK.
$g''(x)
=
-2/x-2e^2/x^3
< 0$.
Finally,
we have something definite!
Since
$g''(x) < 0$
and
$g'(e) = 0$,
$g'(x) > 0$
for $1 < x < e$.
Since
$g'(x) > 0$
and
$g(e) = 0$,
$f'(x)
=g(x) 
< 0$
for $1 < x < e$.
Since
$f'(x) < 0$
and
$f(e) = 0$,
$f(x) > 0$
for
$1 < x < e$
(whew!).
And WE ARE DONE!
A: 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, $x_1x_2 > e^2$.
