Proof of $e^x (\ln x+\frac{1}{x})>\ln 8$ 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 $\ln 8$? 
 A: $f(x)=e^x \left(\dfrac{1}{x}+\log x\right)$
$f'(x)=\dfrac{e^x}{x^2} \left(x^2 \log x+2 x-1\right)$
$f'(x)=0\to x^2 \log x+2 x-1=0\to x\approx 0.59$
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 $x\approx 0.59$

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

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 $\,x>0\,$ .
A: 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
