15. Given the function $f(x)=\frac{4 \mathrm{e}^{x-1}}{x+1}+x^{2}-3 a x+a^{2}-1$, if there exists $x_{0}>0$, such that $f\left(x_{0}\right) \leqslant 0$, find the minimum possible value of the real number $a$.
Solve:
$f(1)=a^{2}-3 a+2=(a-1)(a-2)$, when $a=1$, $f(1)=0$.
Next, we prove that $a>0$ holds for all $(0,+\infty)$.
Since $f(x)=\frac{4 \mathrm{e}^{x-1}}{x+1}+(x-a)^{2}-a x-1 \geqslant \frac{4 \mathrm{e}^{x-1}}{x+1}-a x-1>\frac{4 \mathrm{e}^{x-1}}{x+1}-(x+1)$
$=(x+1)\left[\frac{4 \mathrm{e}^{x-1}}{(x+1)^{2}}-1\right]$, let $g(x)=\frac{4 \mathrm{e}^{x-1}}{(x+1)^{2}}-1, x>0$,
then $g^{\prime}(x)=\frac{4 \mathrm{e}^{x-1}(x+1)-8 \mathrm{e}^{x-1}}{(x+1)^{3}}=\frac{4 \mathrm{e}^{x-1}(x-1)}{(x+1)^{3}}$, thus $g(x)$ is monotonically decreasing on $(0,1)$ and monotonically increasing on $(1,+\infty)$, hence $g(x)_{\min }=g(1)=0 \Rightarrow g(x) \geqslant 0$ always holds.
In summary, the minimum possible value of the real number $a$ is 1.