calculus - 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
$$
1+1/x$$
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.

Answer

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.