Interesting function : $f(x)=\frac{\ln (\frac{x+2}{x+1})}{\ln (1+\frac{1}{x})}$ , $x>1$ Question : 
Prove this function : 

$$f(x)=\dfrac{\ln \left(\dfrac{x+2}{x+1}\right)}{\ln \left(1+\dfrac{1}{x}\right)},\quad x>1$$

is increasing.
I don't know how I solve by my effort is : 
Derivative of $f$ is : 
$$f'(x)=\dfrac{(x+2)\ln \left(\frac{x+2}{x+1}\right)-x\ln \left(\frac{x+1}{x}\right)}{x(x+2)(x+1)\ln^{2} \left(1+\dfrac{1}{x}\right)}$$
Then we will prove that $f'(x)≥0$ for any $x>1$ 
mean that But I don't know how to prove : 

$$(x+2)\ln \left(\frac{x+2}{x+1}\right)-x\ln \left(\frac{x+1}{x}\right)\ge0$$

I need some ideas here if any one have.
Thanks!

I didn't check all the detail of your derivation, but
$$(x+2)\ln \left(\frac{x+2}{x+1}\right)-x\ln \left(\frac{x+1}{x}\right)=\frac{x+2}{x+1}\ln \left(1+\frac1{x+1}\right)^{x+1}-\ln \left(1+\frac1x\right)^x≥0$$
and $$\left(1+\frac1x\right)^x$$ is an increasing function.

EDIT (aimed to clarify any single step)
We start from your last inquality that is

$$(x+2)\ln \left(\frac{x+2}{x+1}\right)-x\ln \left(\frac{x+1}{x}\right)\ge0$$

Now we multiply the LHS by a positive factor $1=\frac{x+1}{x+1}$
$$\frac{x+1}{x+1}(x+2)\ln \left(\frac{x+2}{x+1}\right)-x\ln \left(\frac{x+1}{x}\right)\ge0$$
then we take the upper factor $(x+1)$ at the LHS inside the logarithm using that $n\log A=\log A^n$
$$\frac{x+2}{x+1}\ln \left(\frac{x+2}{x+1}\right)^{x+1}-x\ln \left(\frac{x+1}{x}\right)\ge0$$
similarly, we take also the upper factor $x$ at the RHS inside the logarithm to obtain
$$\frac{x+2}{x+1}\ln \left(\frac{x+2}{x+1}\right)^{x+1}-\ln \left(\frac{x+1}{x}\right)^x\ge0$$
then we use


*

*$\frac{x+2}{x+1}=\frac{1+x+1}{x+1}=1+\frac{1}{x+1}$

*$\frac{1+x}{x}=1+\frac{1}{x}$
to obtain finally

$$\frac{x+2}{x+1}\ln \left(1+\frac1{x+1}\right)^{x+1}-\ln \left(1+\frac1x\right)^x≥0$$

Now we use that


*

*the function $g(x)=\left(1+\frac1x\right)^x$ is (strictly) increasing

*the factor $\frac{x+2}{x+1}>1$
then we can write the latter inequality as

$$\frac{x+2}{x+1}g(x+1)-g(x)\ge \iff \frac{x+2}{x+1}g(x+1)\ge g(x)$$

which is trivially true indeed
$$\frac{x+2}{x+1}g(x+1)>g(x+1)\ge g(x)\quad \blacksquare$$
