Bounding $\ln(1+1/x)$ for $x>0$ I am hoping to somewhat rigorously establish the bound 
$$
\frac{1}{x+1/2}<\ln(1+1/x)<1/x
$$
for any $x>0$. The upper bound is clear since 
$$
1+1/x<e^{1/x} \Rightarrow\ln(1+1/x)<1/x
$$
The lower bound seems dicier. A taylor expansion argument wouldn't seem to work near zero, since $1/x$ blows up. 
Geometrically at least it seems clear near zero. Since $\ln(1+1/x)$ is decreasing and convex, and $\lim_{x\rightarrow 0^+}\ln(1+1/x)=+\infty$ and at 0 
$$
\frac{1}{1/2}=2
$$
I can establish the lower bound near the origin. Should I then take over with taylor expansion for larger $x$? From the graph, it seems like a pretty fine lower bound. 
I still feel like I should be able to make a neater argument using the convexity and decreasing properties of $\ln(1+1/x)$ and any pointers would be appreciated! 
edit: Not a duplicate, the term in the denominator on the lhs has a 1/2.
 A: You can interpret $\ln(x+1/x)=\ln(x+1)-\ln x$ as the area under the graph of $y=1/x$ above $[x,x+1]$.  Since the graph is concave up, the Midpoint rule estimate $M_1=\frac{1}{x+\frac{1}{2}}$ gives an underestimate of the area.
A: Hint: We can write the inequality as
$$e < (1 + 1/x)^{x + 1/2}$$
This quantity approaches $e$ as $x \to \infty$. You're done if you can show that it is strictly decreasing.
A: Let $f(x)=\log\left(1+\frac1x\right)-\frac{1}{x+1/2}$.  
First, note that $\lim_{x\to 0^+}f(x)=\infty$ and $\lim_{x\to \infty}f(x)=0$.  
Moreover, we see that the derivative, $f'(x)$, of $f(x)$ is 
$$\begin{align}
f'(x)&=-\frac{1}{x(x+1)}+\frac{1}{(x+1/2)^2}\\\\
&=\frac{-1/4}{x(x+1)(x+1/2)^2}\\\\
&<0
\end{align}$$
Inasmuch as $f(x)$ is monotonically decreasing to zero, we find that $f(x)>0$ and hence
$$\log\left(1+\frac1x\right)>\frac{1}{x+1/2}$$
as was to be shown!
A: Hint: Define
$$f_1\colon \mathbf R_{>0} \to \mathbf R, \qquad f_1(t):=\ln \left(1+ \frac{1}{t} \right) - \frac{1}{t}$$
and
$$f_2\colon \mathbf R_{>0} \to \mathbf R, \qquad f_2(t):=\ln \left(1+ \frac{1}{t} \right) - \frac{1}{t+\frac{1}{2}}$$
and use the intermediate value theorem twice.
