# 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 $$x_1x_2 > e^2$$

• Here's what I've tried: When $\dfrac{\log x}{x}=a$, $x=e^{ax}.$ Hence, we have: $x_1=e^{ax_1},x_2=e^{ax_2}$. Thus, $x_1x_2=e^{a(x_1+x_2)}$. What have you tried? – user122283 May 24 '14 at 1:12
• @SanathDevalapurkar I am not tried this method. My way is using Cauchy Inequality: Because $x_1x_2 > e^2$ equivalent to $\log (x_1x_2) > 2$. then I have $\log(x_1x_2) = \int_1^{x_1x_2} \frac{1}{t}\text{d}t \geq \left(\int_1^{x_1x_2}\frac{1}{\sqrt{t}}\text{d}t\right)^2 = (2\sqrt{x_1x_2}-2)^2 = 4(\sqrt{x_1x_2}-1)^2>2$. in order to satisfy the last inequality. we must have $x_1x_2 > \left(1+\frac{\sqrt{2}}{2}\right)^2$. the right hand side of the last inequality is less than $e^2$. I hope this way would solve this question.But up to now I can't go any further. – Laura May 24 '14 at 2:01
• $\dfrac{\ln x}x=\dfrac{\ln y}y\iff\sqrt[x]x=\sqrt[y]y\iff x^y=y^x$. – Lucian May 24 '14 at 2:13
• I proved this a number of years ago in sci.math, but I can't find it :( My proof essentially said that $\ln x/x$ gets steeper on the left side of $e$ than on the right. My proof was based on (iirc) the power series for the two curves compared term by term. Someone soon after came up with a better proof involving the derivatives. – marty cohen May 24 '14 at 5:10
• Aha! I got a proof, below. This isn't this one I came up with, but seems to me to be similar to the other one I referred to. – marty cohen May 24 '14 at 5:48

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: $$e^{-1}-a = \int_{t_1}^1 (1-t) e^{-t}\,dt > e^{-1} \int_{t_1}^1 (1-t) \,dt = \frac{(1-t_1)^2}{2e}$$ $$e^{-1}-a = \int_{1}^{t_2} (t-1) e^{-t}\,dt < e^{-1} \int_{1}^{t_2} (t-1) \,dt = \frac{(t_2 - 1)^2}{2e}$$

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!

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, $x_1x_2 > e^2$.

• I give this answer because I'm tickled to have two chances in as many days to make use of the inequality of the geometric and logarithmic means. – user21467 May 24 '14 at 14:02