4. If for any $x \in(0,+\infty)$, the inequality $a \mathrm{e}^{a e x+a} \geqslant \ln (\mathrm{e} x+1)$ always holds, then the range of values for $a$ is $\qquad$
Let $u=\mathrm{e} x+1$, then $a \mathrm{e}^{a u} \geqslant \ln u \Rightarrow a u \mathrm{e}^{a u} \geqslant u \ln u=\ln u \mathrm{e}^{\ln u}$, given $x>0 \Rightarrow u>1 \Rightarrow \ln u>0$, and the function $y=x \mathrm{e}^{x}$ is monotonically increasing on $(0,+\infty)$, thus $a u \geqslant \ln u \Rightarrow a \geqslant \frac{\ln u}{u}$. It is easy to know that the maximum value of the function $f(x)=\frac{\ln x}{x}$ is $f(\mathrm{e})=\frac{1}{\mathrm{e}}$, so the range of $a$ is $\left[\frac{1}{\mathrm{e}},+\infty\right)$.