Prove the inequality $\ln {(1+\frac{1}{x})}> \frac{2}{2x+1}$ Prove the inequality $$\ln {(1+\frac{1}{x})}> \frac{2}{2x+1}$$
for $x>0$.
My attempt: Let $$f(x)=\ln {(1+\frac{1}{x})}-\frac{2}{2x+1}$$
Then $$f'(x)=-\frac{1}{x(x+1)}+\frac{4}{(2x+1)^2}$$
$$f''(x)=\frac{1}{x^2}-\frac{1}{(x+1)^2}-\frac{8}{(2x+1)^3}>0$$
Then the function $f$ is convex. There exists a minimal point $x_0$ such that $f(x)\geq f(x_0)$. However, there's no critical point $x_0$ such that $f'(x_0)=0$, and $\lim_{x \rightarrow \infty} \sup {f'(x)}=0$. Then I want to show that $f(x)>0$, how do I continue my proof?
I have been trying another approach using Cauchy's MVT by letting
$$f(x)=\ln {x}$$ $$g(x)=\frac{1}{2x+1}$$ such that $$\frac{f(x+1)-f(x)}{g(x+1)-g(x)}=\frac{f'(c)}{g'(c)}$$ where $c \in (x,x+1)$ but failed. 
As what I did is
$$\ln {(1+\frac{1}{x})}=\frac{1}{c} \cdot \frac{(2c+1)^2}{2} \cdot \frac {2}{(2x+1)(2x+3)}$$
I can't simply do the inequality
$$\frac{1}{c} \cdot \frac{(2c+1)^2}{2}>\frac{1}{x} \cdot \frac{(2x+1)^2}{2}$$
as $c>x$ because $\frac{1}{c} < \frac{1}{x}$ but $\frac{(2c+1)^2}{2} > \frac{(2x+1)^2}{2}$.
Edited: Of course, I know that $$\ln {(1+ \frac{1}{x})}>\frac{x}{1+x}$$
for $x>-1$. I just need to prove that $$\frac{x}{x+1}>\frac{2}{2x+1}$$ But I hope to find out another approach using calculus method.
 A: With your approach:
$$
f'(x)=-\frac{1}{x(x+1)}+\frac{4}{(2x+1)^2} = \frac{-1}{x(x+1)(2x+1)^2} < 0
$$
so that $f$ is strictly decreasing, and therefore
$$
 f(x) > \lim_{t\to \infty} f(t) = 0 \, .
$$
Or you substitute $y = 1/x$ and consider
$$
 g(y) = \ln (1+y) - \frac{2y}{2+y}
$$
with $g(0) = 0$ and
$$
 g'(y) = \frac{y^2}{(y+1)(y+2)^2} > 0 \, .
$$

Remark: The estimate $\ln {(1+ \frac{1}{x})}>\frac{x}{1+x}$ is not good enough because   $\frac{x}{x+1}<\frac{2}{2x+1}$ for small $x$.
A: Here is a more natural way to prove this. Note that:
$$\ln\bigg(1+\frac{1}{u}\bigg) = \int\limits_u^{u+1} \frac{1}{x} dx$$
i.e. the area under the graph of $\frac{1}{x}$ between $u$ and $(u+1)$. This explains why $\frac{2}{2u+1}$ is such a good approximation in the first place -- it comes from approximating the area as a rectangle of width $1$, and height $f\big(\big(u+\frac{1}{2}\big)\big)=\frac{1}{\big(u+\frac{1}{2}\big)}.$
Let $v = \big(u+\frac{1}{2}\big)$ for convenience. Then, an easy way to show that $\frac{2}{2u+1} = \frac{1}{v}$ underestimates the area is to show that:
$$\Bigg( \frac{1}{v-h} - \frac{1}{v} \Bigg) \ge \Bigg( \frac{1}{v} - \frac{1}{v+h} \Bigg)$$
where $h > 0$.
This is clearly true. Thus, the inequality holds.
