Proof of $e^x (\ln x+\frac{1}{x})>\ln 8$

Question

Prove that $$e^x \left(\ln x+\frac{1}{x}\right)>\ln 8$$

I found that the minimum of $e^x \left(\ln x+\frac{1}{x}\right)$ is close to $\ln 8$, then how do we prove that it's greater than $\ln 8$?

Answer 1

$f(x)=e^x \left(\dfrac{1}{x}+\log x\right)$

$f'(x)=\dfrac{e^x}{x^2} \left(x^2 \log x+2 x-1\right)$

$f'(x)=0\to x^2 \log x+2 x-1=0\to x\approx 0.59$

Taylor polynomial at $x=\dfrac{1}{2}$

$f(x)=\sqrt e \left[x^2 \left(5 -\dfrac{1}{8} \log 16\right)+x \left(-5 -\dfrac{1}{8} \log 16\right)+\dfrac{13 }{4}-\dfrac{1}{8} \log 32\right]+O(x^3)$

$f(x)\approx 7.6722 x^2-8.81501 x+4.64409$ in a neighbourhood of $x=\frac12$

To estimate the error we need the third derivative

$f^{(3)}(x)=\dfrac{e^x}{x^4} \left(x^4 \log (x)+4 x^3-6 x^2+8 x-6\right)$

On the interval $[0.4,0.6]$ we have $|f^{(3)}(x)|\le 36.0234$

thus the error is $R_3(x)\le \dfrac{ |f^{(3)}(x)| \cdot \left|\,x-\dfrac{1}{2}\right|}{3!}\approx 0.006$

Now as $7.6722 x^2-8.81501 x+4.64409>\log 8;\;\forall x\in\mathbb{R}$

we can conclude that $f(x)>\log 8$ for any $x\in\mathbb{R}$

Hope this helps

Edit

A graph can explain better. Remember that the minimum is at $x\approx 0.59$

Answer 2

Rewrite the inequality as $$f(x) = \ln(x) + \frac{1}{x} > \frac{\ln8}{e^x}=g(x).$$ $$f(x) > g(x)$$ The derivatives are: $$f'(x) = \frac{\partial f(x)}{\partial x} = \frac{x-1}{x^2},$$ $$g'(x) = \frac{\partial g(x)}{\partial x} = -\frac{\ln8}{e^{x}}<0. $$ Clearly, $f'(x)$ is negative for $x\in [0,1)$ and positive for $x\geq 1$. It is easy to verify that $f(x)$ obtains its minimum at $x=1$, which is equal to $f(1)=1$.

Observe that $g(x=1) = \frac{\ln 8 }{e} <1$, which means that for $x \geq 1$, it is clearly $f(x) > g(x)$.

Now, for $x \in [0,1)$, first observe that

$$\lim_{x \to 0} f(x) = \lim_{x \to 0} \left(\ln(x) + \frac{1}{x}\right) = \infty,$$

which follows from the fact that, as $x$ decreases from $1$ to $0$, the rate of decrease of the first term is lower than the rate of increase of the second term. Also observe that $$\lim_{x \to 0} g(x) = \lim_{x \to 0} \frac{\ln 8}{e^x} = \ln 8 < \infty.$$

This means that if $g(x)>f(x)$ for $x \in [a,b]$ (where $0<a\leq b<1$), then $g'(x)=f'(x)$ for some $x\in[a,b]$. Suppose this is the case, then $g'(x)=f'(x)$ means

$$\frac{1-x}{x^2} = \frac{\ln 8}{e^x}.$$

Note that if there is a solution to the above equation, this solution is unique as the RHS is increasing in $x$ while the LHS is decreasing in $x$ (it easy to verify). Denote the solution by $x^*.$ This in its turn implies that the following must be true:

$$\ln(x^*) + \frac{1}{x^*} < \frac{\ln8}{e^{x^*}} = \frac{1-x^*}{(x^*)^2},$$

which means that $f(x^*)<-f'(x^*)$. For this inequality to be satisfied, we know that $-f'(x^*) >1$, or \begin{equation} x<\frac{\sqrt{5}-1}{2} \tag{eq.1} \end{equation} However, notice that at $x=2/3 (>\frac{\sqrt{5}-1}{2})$, we have $$-f'(x) <- g'(x)$$ which implies that $x^*$ must be greater than $2/3$, which contradicts to (eq.1).

Answer 3

$x>0$ : $\enspace\displaystyle e^{-x}x\ln 8 < 0.39+\frac{x}{2} < 1 + x\ln x$

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Claim: $\enspace\displaystyle e^x (\ln x + \frac{1}{x}) > \ln 8 \,$ or equivalent $\,\displaystyle e^{-x}x\ln 8 < 1+x\ln x \,$ for $\,x>0\,$

We are looking for $\,a\,$ and $\,b\,$ with: $\,\text{(A)}$ $\enspace\displaystyle e^{-x}x\ln 8 \leq a+\frac{x}{2}\enspace$ and $\,\text{(B)}$ $\enspace\displaystyle b+\frac{x}{2} \leq 1+x\ln x \,$

$\text{(A)}$ $\,$ Be $\,W(x)\,$ the main branch of the Lambert W-function: $\,W(x)e^{W(x)}=x$ with $\,\displaystyle x>-\frac{1}{e}$

Minimum of $\,\displaystyle a+\frac{x}{2} - e^{-x}x\ln 8\,$ :

$\displaystyle (a+\frac{x}{2} - e^{-x}x\ln 8)’=\frac{1}{2}+(x-1) e^{-x}\ln 8 := 0\,$ or equivalent $\,\displaystyle (1-x)e^{1-x}=\frac{e}{2\ln 8}$

=> $\enspace\displaystyle x=1-W(\frac{e}{2\ln 8}) $

It follows the minimum with

$\displaystyle (a+\frac{x}{2} - e^{-x}x\ln 8)’’|_{ x=1-W(\frac{e}{2\ln 8}) }=\frac{1}{2}(1+1/ W(\frac{e}{2\ln 8}))\approx 1.672… > 0$

and the calculation of $\,a\,$ with

$\displaystyle (a+\frac{x}{2} - e^{-x}x\ln 8)|_{ x=1-W(\frac{e}{2\ln 8})}=a+1-\frac{x}{2}( W(\frac{e}{2\ln 8})+1/W(\frac{e}{2\ln 8}))\geq 0$

which gives the $a$-minimum value

$\displaystyle a:= -1+\frac{1}{2}( W(\frac{e}{2\ln 8})+1/W(\frac{e}{2\ln 8}))\approx 0.385317…<0.39\,$ .

$\text{(B)}$ $\,$ Same way as before, but much easier.

Minimum of $\,\displaystyle 1-b+x\ln x-\frac{x}{2} \,$ :

$\displaystyle (1-b+x\ln x-\frac{x}{2})’=\frac{1}{2}+\ln x := 0\enspace$ => $\enspace\displaystyle x=\frac{1}{\sqrt{e}}$

It follows the minimum with $\enspace\displaystyle (1-b+x\ln x-\frac{x}{2})’’|_{ x=\frac{1}{\sqrt{e}} }=\sqrt{e}>0 $

and the calculation of $\,b\,$ with $\enspace\displaystyle (1-b+x\ln x-\frac{x}{2})|_{ x=\frac{1}{\sqrt{e}} }=1-b-\frac{1}{\sqrt{e}}\geq 0$

which gives the $b$-maximum value $\,\displaystyle b:= 1-\frac{1}{\sqrt{e}}\approx 0.393469…>0.39\,$ .

Result: $\,$ The claim is correct since $\enspace\displaystyle e^{-x}x\ln 8 < 0.39+\frac{x}{2} < 1 + x\ln x \,$ for $\,x>0\,$ .

Answer 4

The function $$f(x)=\frac{e^x}{x}(x\log x+1)-\log 8$$ is defined for $x >0$ and we want to show that $$f(x)>0, \forall x >0.$$

We want to prove this and the proof will involve floating point calculations. Nevertheless the proof should be logically correct.

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The expressions we calculate are composed of constants, addition, subtraction, multiplication, division, powers and evaluation of elementary transzendental functions (exp, log).

If $\text{expr}$ is such an arithmetic expression then we write $\#[\text{expr}]$ for the numerical value calculated. In the table below the values have a precision of three digits after the decimal point. This means that $|\text{expr} - \overline{\text{expr}}|<\delta_0,$ where $\delta_0=0.0005$.

$$ \begin{array}{ r | r | r |l|r } \text{line} & x, x_1& x_2&\text{expr} & \#[\text{expr}] \\ \hline 1&&&-2e^{-\frac{3}{2}}+2& 1.554\\ \hline 2&0.591&&x^2 \log x+2x-1& -0.002\\ 3&0.593&&x^2 \log x+2x-1& 0.002\\ 4&0.591&0.593& {e^{x_1}}\left(\log x_1+\dfrac{1}{x_2}\right)-\log 8&0.016\\ 5&0.591&0.593&{e^{x_1}}\left(\log x_1+\dfrac{1}{x_1}\right)-\log 8 +\\ &&&\qquad{e^{x_1}}\left( \log x_1 +\dfrac{2}{x_1}-\dfrac{1}{x_1^2}\right)(x_2-x_1)&0.026\\ \end{array} $$

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We plot $f$ in a subinterval:

$f$ seems to be a convex function with a global minimum at $x_0 \approx 0.6.$

If so, the first derivative $f’$ then must be an increasing function with a zero at $x_0,$ and the second derivative $f’’$ must be a positive function.

$$f'(x)=\frac{e^x}{x^2}(x^2 \log x +2x-1)$$ $f’$ has a zero at $x_0.$

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The proof that $f(x)>0,\; \forall x>0,$ consists of two steps:

1. prove that $f$ has a global minimum at $x_0$
2. prove that $f(x_0)>0$

The first step is more or less identical to the calculations in this answer. We repeat it here for the sake of completeness.

We have $$f'(x)=\frac{e^x}{x^2}h(x)$$ where $$\begin{eqnarray} h(x)&=&x^2 \log x+2x-1\\ h'(x)&=&2x \log x+x+2\\ h''(x)&=&2\log x+3 \end{eqnarray}$$

We want to show that $$h'(x)>0, \; x>0$$ This is a problem similar to our original problem.

$h''$ is an increasing function with a zero at $e^{-\frac{3}{2}}$, so $h'$ has a global minimum at $e^{-\frac{3}{2}}$ and

$$h’(e^{-\frac{3}{2}}) =-2e^{-\frac{3}{2}}+2 \ge \#[-2e^{-\frac{3}{2}}+2] - \delta_0 \ge1.5535 >0, \tag{1}$$

Therefore $h$ is an increasing function and we have $$h(0.591)\le \#[h(0.591)]+\delta_0<0<\#[h(0.593)]-\delta_0<h(0.593)$$

Because $h$ is increasing it has exactly one root $x_0$ and this is in $(0.591, 0.592)$

So we have $$h(x) \begin{cases} <0, \;0<x< x_0\\ =0,\; x=x_0 \\ >0,\;x>x_0 \end{cases}$$ and $x_0 \in (0.591, 0.592)$. The same is true for $f’$ because it is a positive multiple of $h$ and so $f$ has a global minimum at $x_0$. This completes the first part of our proof.

Now we have to prove that $f(x_0)>0$ for this $x_0 \in (x_1, x_2)$. It does not help to plug in the approximation into the function but we can do the following estimation:

$$\begin{array}\\ \tag{2} f(x_0)&=&e^{x_0}(\log x_0+\dfrac{1}{x_0})-\log 8\\ &\gt& e^{x_1}(\log x_1+\frac{1}{x_2})-\log 8\\ &\ge& \#[e^{x_1}(\log x_1+\frac{1}{x_2})-\log 8]-\delta_0\\ &=&0.0155\\ &>&0 \end{array}$$

Or we can use the Mean Value Theorem

$$\exists \xi \in (x_1, x_0):\; \frac{f(x_0)-f(x_1)}{x_0-x1}=f'(\xi)$$

and therefore ($f'$ increasing, $f'(x_1)<0$) $$\begin{array}\\ \tag{3} f(x_0)&=&f(x1)+f’(\xi)(x_0-x_1)\\ &\ge& f(x1)+f’(x_1)(x_2-x_1) \\ &=&{e^{x_1}}\left(\log x_1+\dfrac{1}{x_2}\right)-\log 8 +{e^{x_1}}\left( \log x_1 +\dfrac{2}{x_1}-\dfrac{1}{x_1^2}\right)(x_2-x_1)\\ &\ge& \#[{e^{x_1}}\left(\log x_1+\dfrac{1}{x_2}\right)-\log 8 +{e^{x_1}}\left( \log x_1 +\dfrac{2}{x_1}-\dfrac{1}{x_1^2}\right)(x_2-x_1)]-\delta_0\\ &=&0.0255\\ &>&0 \end{array}$$

Answer 5

I got a new way to calculate the minimum value. Let $$\begin{eqnarray} f(x)&=&e^x(\ln x+\frac{1}{x})\\ f'(x)&=&\frac{e^x}{x^2}h(x)\\ h(x)&=&x^2 \ln x+2x-1\\ h'(x)&=&2x \ln x+x+2\\ h''(x)&=&2\ln x+3 \end{eqnarray}$$

• $h^{"}(x)$ has a zero point at $x=e^{-\frac{3}{2}}$,
• when $x>e^{-\frac{3}{2}}$ , $h^{"}(x)>0$,
• when $x<e^{-\frac{3}{2}}$ , $h^{"}(x)<0$

Therefore $h^{'}(x)>h^{'}(e^{-\frac{3}{2}})>0$ and $h(x)$ is increasing for $x>0$.

$h(1/e)<0$ and $h(0.594)=0.00421621...>0$ therefore exist $x_0 \in(1/e,0.594)$ such that $h(x_0)=0$, so \begin{equation} x_0^2\ln x_0+2x_0-1=0 \tag{1} \end{equation}

$h(x)$ is increasing on $x>0$ $$\begin{cases} h(x)<0,x\in(0, x_0)\\ h(x)>0,x\in(x_0,+\infty) \tag{2} \end{cases}$$ Then we have $$\begin{cases} f^{'}(x)<0,x\in(0,x_0)\\ f^{'}(x)>0,x\in(x_0,+\infty) \tag{3} \end{cases}$$ thus $$\min f(x)=f(x_0)=e^{x_0}(\ln x_0+\frac{1}{x_0})$$ from equation $(1)$ we have $\min f(x)=e^{x_0}\frac{1-x_0}{x_0^2}$

Let $$g(x)=e^x \frac{1-x}{x^2}$$ then $g^{'}(x)<0$ for $x>0$. Because $x_0 \in (1/e,0.594)$ we have $$g(x_0)>g(0.594)=2.08413...>\ln 8.$$ and further $$\min f(x)>\ln 8$$

Q.E.D