$ a^x ≥ x +1, x∈R $
What is the range of x such that this inequation is constantly true？
I tried to transform the inequation into the following：
$
xlna ≥ ln(x+1)
$, which can get the following inequation
$$\begin{cases} \ lna ≥ \frac{ln(x+1)}{x}, x＞ 0\\ \ lna ≤ \frac{ln(x+1)}{x}, x < 0 \end{cases}$$
In two cases, it seems we cannot calculate the maximum value of the right-side equation, which in turn, determines the range of a.
The derivative to the right-side equation is
$$\begin{align} \frac{\frac{1}{x+1}-{ln(x+1)}}{x^2} \end{align}$$
It seems that we cannot figure out the extreme point
