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!
 A: 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
$$\frac{x+2}{x+1}g(x+1)>g(x+1)\ge g(x)\quad \blacksquare$$
A: EDIT
Here's a proof without expansions.
We wish to prove that 
$$f(x) = \frac{\log (\frac{x+2}{x+1})}{\log (\frac{x+1}{x})}$$
is an increasing function, i.e. that 
$$f'(x) = \frac{(x+2) \log \left(\frac{1}{x+1}+1\right)-x \log \left(\frac{1}{x}+1\right)}{x (x+1) (x+2) \log ^2\left(\frac{1}{x}+1\right)}>0$$
Since the denominator is positive we need to show that the numerator is positive, i.e.
$$(x+2) \log \left(\frac{1}{x+1}+1\right)-x \log \left(\frac{1}{x}+1\right)>0\tag{1}$$
Now the l.h.s. of $(1)$ can easily be shown to be identical to the integral
$$g= \int_x^{x+1} \log \left(\frac{1}{t}+1\right) \, dt\tag{2}$$
which is a positive quantity since the integrand is positive.
This completes the proof. Q.E.D.
Corollary
Using the same technique we can easily prove that 
$$h(x) = \left(1+\frac{1}{x}\right)^x$$
a strictly increasing function for $x>0$.
Indeed, the derivative is
$$h'(x) = \frac{\left(\frac{1}{x}+1\right)^x \left((x+1) \log \left(\frac{1}{x}+1\right)-1\right)}{x+1}$$
Hence we need to show that
$$\log \left(\frac{1}{x}+1\right)-\frac{1}{x+1}>0 $$
But this is obvious since the l.h.s. is equal to the positive integral
$$\int_x^{\infty } \frac{1}{t (t+1)^2} \, dt$$
Original post
For $x\gt0$ we have
$$f(x) = \frac{\log (\frac{x+2}{x+1})}{\log (\frac{x+1}{x})}\\=\frac{\log (\frac{x+1+1}{x+1})}{\log (\frac{x+1}{x})}\\=\frac{\log \left(1+\frac{1}{x+1}\right)}{\log \left(1+\frac{1}{x}\right)}\\\simeq \frac{{\frac{1}{x+1}}}{\frac{1}{x}}=\frac{x}{x+1}=1-\frac{1}{x+1}$$
In the last line we have assumed that $x>>1$ and have used the expansion for small $|\epsilon|<<1$
$$\log(1+\epsilon)= \epsilon-\frac{1}{2}\epsilon^2+ ...$$
Hence
$$\frac{d}{dx}f(x)\simeq \frac{d}{dx}(1-\frac{1}{x+1})= +\frac{1}{(x+1)^2}) \gt 0$$
showing that the function is increasing. Q.E.D.
