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 $\ln 8$? 

$f(x)=e^x \left(\dfrac{1}{x}+\log x\right)$
$f'(x)=\dfrac{e^x}{x^2} \left(x^2 \log x+2 x-1\right)$
$f'(x)=0\to x^2 \log x+2 x-1=0\to x\approx 0.59$
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 $x\approx 0.59$

