Prove that $0 < g(x) - \ln x \leq \ln\left(\frac{x+2}{x}\right)$ for $x > 0$ $$g(x) = \ln(x + 1 + e^{-x})$$
My question is prove that $0 < g(x) - \ln x \leq \ln\left(\frac{x+2}{x}\right)$ for $x > 0$
How do I do that?
My attempts:
I have only successfully proved the second part:
We know that $x > 0$ then
$e^{-x} + x + 1 < 2 + x$
$1/x$ is positive so we multiply each other by $1/x$ then we put $\ln$ on each side too, we get this
$\ln (\frac{e^{-x} + x + 1}{x}) < \ln\left(\frac{2 + x}{x}\right)$
And since $g(x) - \ln x$ can be written as $\ln \left(\frac{e^{-x} + x + 1}{x}\right)$ then we proved it.
We get $g(x) - \ln x < \ln\left(\frac{2 + x}{x}\right)$ which is the second part.
The first part I have tried to prove it but I never succeeded:
$x > 0$
$x + 1 > 1$
$e^{-x} < 1$
Their signs are different so no idea what to do after this. Any idea?
 A: \begin{align}
& \ln(x+1+e^{-x}) - \ln x \\[8pt]
= {} & \ln\frac{x+1+e^{-x}} x \\[8pt]
\le {} & \ln\frac{x+1+1} x
\end{align}
A: The function $\ln(x)$ is strictly increasing. Then, $\ln(x) < \ln(x+1+e^{-x})$ holds iff
$x < x+1+e^{-x}$, i.e. iff $-1 < e^{-x}$, and this inequality holds for every $x>0$ since the exponential function is positive (indeed it holds for every real number, but here you can just consider $x>0$ because $x$ should be in the domain of the logarithmic function).

Edit (03-05-2021): Trying to explain more carefully the argument:
The facts I suppose you know about the functions $g(z)=ln(z)$ and $h(z)=e^z$ are

*

*The function $g(z)=ln(z)$ is strictly monotonically increasing; that means that $g(z_1) < g(z_2) $ if and only if $z_1 < z_2$;

*The function $h(z)=e^z$ can only take positive values (indeed, it's a bijection between $R$ and $R^+$).

With those points accepted, let's prove that
$$\forall x>0: ln(x)<ln(x+1+e^{-x}).$$
By the first point, if we show that
$$\forall x>0: x<x+1+e^{-x},$$
we will be done, since the both statements above are equivalent (just replace $z_1$ by $x$ and $z_2$ by $x+1+e^{-x}$).
But we can rewrite our last statement as
$$\forall x>0: -1<e^{-x},$$
(just add $-x-1$ in both sides of the inequality). Then, your initial problem was reduced to proving that $\forall x>0: -1<e^{-x}$, but by our second point this is trivially true.
