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.
 A: 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 $f'(x)>0$.
A: 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
$$\frac{1}{t+|x|}>\frac{1}{1+|x|}$$
A: 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:
$$\lim_{x\to -\infty}f(x)=0^+$$
