Monotonicity of the function $(1+x)^{\frac{1}{x}}\left(1+\frac{1}{x}\right)^x$. Let $f(x)=(1+x)^{\frac{1}{x}}\left(1+\frac{1}{x}\right)^x, 0<x\leq 1.$
Prove that $f$ is strictly increasing and $e<f(x)\leq 4.$
In order to study the Monotonicity of $f$, let 
$$g(x)=\log f(x)=\frac{1}{x}\log (1+x)+x\log \left(1+\frac{1}{x}\right).$$
And $f$ and $g$ has the same Monotonicity. By computation,
$$g'(x)=\frac{1}{x^2}\left(\frac{x}{1+x}-\log (1+x)\right)+\log \left(1+\frac{1}{x}\right)-\frac{1}{1+x}.$$ 
As we know $\frac{x}{1+x}-\log (1+x)\leq 0$ and $\log \left(1+\frac{1}{x}\right)-\frac{1}{1+x}\geq 0$. So it does not determine the sign of $g'(x)$.
If we compute the second derivative $g''(x)$,  you will find it is also difficult to determine the sign of $g''(x)$.
 Our main goal is to prove $$\frac{1}{x^2}\left(\frac{x}{1+x}-\log (1+x)\right)+\log \left(1+\frac{1}{x}\right)-\frac{1}{1+x}>0.$$
Is there some tricks to prove this result. Any help and hint will welcome.

Michael Rozenberg's answer has it all. 
Here are two remaining proofs in Michael Rozenberg's answer: 


*

*Show: $\ln{x}\leq\frac{2(x-1)}{1+x}$


We have $\ln(1+y)\leq\frac{2y}{{2+y}}$ for $-1<y<0$ (see e.g. in https://en.wikipedia.org/wiki/List_of_logarithmic_identities#Inequalities), 
i.e. $\ln(x)\leq\frac{2(x-1)}{{1+x}}$ for $0<x<1$. This is exactly what needs to be shown.


*Show: $\ln(1+x)\leq\frac{x(2x+1)}{(1+x)^2}$


We have $\ln(1+x)\leq\frac{x}{\sqrt{1+x}}$ (see e.g. in https://en.wikipedia.org/wiki/List_of_logarithmic_identities#Inequalities), so we may prove
$\frac{1}{\sqrt{1+x}}\leq\frac{2x+1}{(1+x)^2}$ or $(1+x)^3-{(1+2x)^2}\leq 0$ or $x^2 - x - 1<0$ or $x( 1-x) + 1>0$ which is true since $1-x \ge 0$.
