15. (15 points) Given that when $x \in[1, \mathrm{e}]$, the inequality
$$
a \ln x \leqslant-\frac{1}{2} x^{2}+(a+1) x
$$

always holds. Try to find the range of the real number $a$.
Three, 15. The inequality can be transformed into
$$
a(x-\ln x) \geqslant \frac{x^{2}}{2}-x \text {. }
$$

Since $x \in[1, \mathrm{e}]$, we have $x-\ln x>0$.
Thus, the inequality becomes $a \geqslant \frac{\frac{x^{2}}{2}-x}{x-\ln x}$.
Let $g(x)=\frac{\frac{x^{2}}{2}-x}{x-\ln x}(x \in[1, \mathrm{e}])$. Note that
$$
g^{\prime}(x)=\frac{(x-1)\left(\frac{x}{2}+1-\ln x\right)}{(x-\ln x)^{2}}>0,
$$

where $x \in(1, \mathrm{e})$, and $g(x)$ is continuous at $x=1$ and $x=\mathrm{e}$, so $g(x)$ is an increasing function on $x \in[1, \mathrm{e}]$.
Therefore, $a \geqslant g(\mathrm{e})=\frac{\mathrm{e}^{2}-2 \mathrm{e}}{2(\mathrm{e}-1)}$.