Show that, for all $x>0: \frac{1}{x+1/2} \leq \ln(1+1/x)$

Question

I was able to prove that $\ln(1+1/x) \leq 1/x$ by integrating both sides and taking the upper boundaries of the integral area. However, that does not seen to work for the inequation in question. Any ideas? Thanks.

Answer 1

The function $t \mapsto 1/t$ is convex for $t \geq 1$. Therefore it lies above any of its tangent lines on this interval. Use this to get a lower bound for $\log(1+x)$ for any $x\geq 0$:

$$\log(1+x) = \int_1^{1+x} \frac{\mathrm dt}{t} \geq \int_1^{1+x} \ell(t) \mathrm dt$$

where $t \mapsto \ell(t)$ is the tangent line of $t \mapsto 1/t$ at $t = 1+x/2$. This results in $$\log(1+x) \geq \frac{2x}{x+2}$$ for any $x\geq 0$. Now replace $x$ by $1/x$.