Proof of $e^x (\ln x+\frac{1}{x})>\ln 8$ Prove that 
$$e^x \left(\ln x+\frac{1}{x}\right)&gt;\ln 8$$

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 $\ln 8$?
$f(x)=e^x \left(\dfrac{1}{x}+\log x\right)$

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

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

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 
$7.6722 x^2-8.81501 x+4.64409&gt;\log 8;\;\forall x\in\mathbb{R}$

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

Hope this helps

Edit

A graph can explain better. Remember that the minimum is at $x\approx 0.59$
I got a new way to calculate the minimum value.
Let
$$\begin{eqnarray}
f(x)&amp;=&amp;e^x(\ln x+\frac{1}{x})\\
f'(x)&amp;=&amp;\frac{e^x}{x^2}h(x)\\
h(x)&amp;=&amp;x^2 \ln x+2x-1\\
h'(x)&amp;=&amp;2x \ln x+x+2\\ 
h''(x)&amp;=&amp;2\ln x+3
\end{eqnarray}$$


$h^{"}(x)$ has a zero point at $x=e^{-\frac{3}{2}}$, 
when $x&gt;e^{-\frac{3}{2}}$ , $h^{"}(x)&gt;0$, 
when $x&lt;e^{-\frac{3}{2}}$ , $h^{"}(x)&lt;0$


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

$h(1/e)&lt;0$  and $h(0.594)=0.00421621...&gt;0$ 
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 $x&gt;0$
$$\begin{cases} h(x)&lt;0,x\in(0, x_0)\\ h(x)&gt;0,x\in(x_0,+\infty) \tag{2} \end{cases}$$
Then we have
$$\begin{cases} f^{'}(x)&lt;0,x\in(0,x_0)\\ f^{'}(x)&gt;0,x\in(x_0,+\infty) \tag{3} \end{cases}$$
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 $g^{'}(x)&lt;0$ for $x&gt;0$.
Because $x_0 \in (1/e,0.594)$
we have 
$$g(x_0)&gt;g(0.594)=2.08413...&gt;\ln 8.$$
and further
$$\min f(x)&gt;\ln 8$$

Q.E.D