Prove this inequality $(2+x)\ln{x}-e(x-1)>0,x>1$ 
Let $x>1$,show that
  $$(2+x)\ln{x}-e(x-1)>0$$

The simaler problem:[How prove $(2+5x)\ln{x}-6(x-1)>0.\forall x>1$ 
this problem idea:  $f(x)=\ln{x}-\dfrac{e(x-1)}{x+2}$
then
$$f'(x)=\dfrac{x^2+(4-3e)x+4}{x(x+2)^2}=\dfrac{\left(x-\dfrac{3e-4}{2}\right)^2+4-\dfrac{(3e-4)^2}{4}}{x(x+2)^2}$$
but $f'(x)>0,\forall x>1$ is not  always hold,because
$$4-\dfrac{(3e-4)^2}{4}<0$$

It's easier to deal with $f(x)=(2+x)\ln x-e(x-1)$ directly. Note that $f(1)=0$.
Since $f'(x)=\ln x+\frac2{x}+1-e$ and $f''(x)=\frac{x-2}{x^2}$, $f'(x)$ is monotone decreasing in $1<x<2$ and monotone increasing in $x>2$.
From $f'(2)=\ln 2+2-e<0$ and $f'(2.6)=\ln 2.6+\frac2{2.6}+1-e>0$, there is a constant $2<c<2.6$ for which $f'(c)=0$. (Finding $2.6$ is a bit tricky. I used WolframAlpha.)
Then you can show that 
\begin{align*}
f(c)&=(2+c)\ln c-e(c-1)\\
&=(2+c)(-\frac2c-1+e)-e(c-1)\\
&=-c-\frac4c+3e-4\\
&>-2.6-\frac4{2.6}+3e-4\\
&>0.\ \text{(Very subtle.)}
\end{align*}
The above argument is sufficient to show $f(x)>0$ for $1<x<2$, $2\le x<2.6$ and $2.6\le x$.

