Prove that $\ln\left(1 + \frac{1}{|x|}\right) - \frac{1}{1 + |x|}$ is always positive

I have to study when the function $$f'(x) = \ln\left(1 + \frac{1}{|x|}\right) - \frac{1}{1 + |x|}$$ is positive. I tried to use the inequality $\ln(1 + t) < t$, $\forall t > -1$ but this could not help me ($\frac{1}{|x|} - \frac{1}{1 + |x|}$ is always a positive quantity but I cannot say anything about $f'(x)$).

Is there a quick method? Thanks.

Similarly to $\ln(1+t)\leq t$, we have a well-known inequality in the other direction: $$\ln(1+t)\ge \frac{t}{1+t}\quad\text{for }t>-1 $$ with equality only if $t=0$. Plugging in $t=\frac{1}{|x|}\ne 0$ will give you $f'(x)>0$.

Fix $x\neq 0$. For each $x\neq 0$, apply the mean value theorem to the function

$$f(u):=\ln{(u+|x|)}$$ on $[0,1]$. You get

$$\frac{\ln{(1+|x|)}-\ln{|x|}}{1-0}=\frac{1}{t+|x|}$$ for some $0<t<1$. Notice that $$\ln{(1+|x|)}-\ln{|x|}=\ln{\left(1+\frac{1}{|x|}\right)}$$ and that $$\frac{1}{t+|x|}>\frac{1}{1+|x|}$$

The function is simmetric in rispect to the $y$ axis. We can, thus, study the first derivate for $x>0$. We have: $$f'(x)=-\frac{1}{1+\frac{1}{x}}\cdot\frac{1}{x^2}+\frac{1}{(x+1)^2}=-\frac{1}{x\cdot (1+x)}+\frac{1}{(1+x)^2}<0\,\,\,\forall x \in \mathbb{D}_{f'(x)}$$

Also: $$\lim_{x\to 0^+}f(x)=+\infty$$ And: $$\lim_{x\to +\infty}f(x)=0^+$$

So, in $(0,+\infty)$ the function is always positive and decreasing. In $(-\infty, 0)$, $f(x)$ is again always positive and increasing. Namely: $$\lim_{x\to -\infty}f(x)=0^+$$