14. Given $f(x)=\mathrm{e}^{x}$.
(1) When $x \geqslant 0$, the inequality $(x-1) f(x) \geqslant m x^{2}-1$ always holds, find the range of $m$;
(2) Prove that when $x>0$, $f(x)>4 \ln x+8-8 \ln 2$.
(1) $(x-1) \mathrm{e}^{x}+1 \geqslant m x^{2}, x=0$ when, $m \in \mathbf{R}$; $x>0$ when, $m \leqslant \frac{(x-1) \mathrm{e}^{x}+1}{x^{2}}$,

Let $g(x)=\frac{(x-1) \mathrm{e}^{x}+1}{x^{2}}, g^{\prime}(x)=\frac{x^{2} \mathrm{e}^{x}-2\left[(x-1) \mathrm{e}^{x}+1\right]}{x^{3}}=\frac{\left(x^{2}-2 x+2\right) \mathrm{e}^{x}-2}{x^{3}}$, 
Let $h(x)=\left(x^{2}-2 x+2\right) \mathrm{e}^{x}-2, h^{\prime}(x)=\left(x^{2}+2\right) \mathrm{e}^{x}>0 \Rightarrow h(x)>h(0)=0$, 
Thus $g^{\prime}(x)>0 \Rightarrow g(x)>\lim _{x \rightarrow 0} \frac{(x-1) \mathrm{e}^{x}+1}{x^{2}}=\lim _{x \rightarrow 0} \frac{\mathrm{e}^{x}}{2}=\frac{1}{2}$.
Therefore, the range of $m$ is $\left(-\infty, \frac{1}{2}\right]$.
(2) Let $p(x)=\mathrm{e}^{x}-4 \ln x-8+8 \ln 2, p^{\prime}(x)=\mathrm{e}^{x}-\frac{4}{x}$, since $p^{\prime}(x)$ is monotonically increasing on $(0,+\infty)$,
and $p^{\prime}(1)=\mathrm{e}-4<0$, there exists $x_{0} \in(1,2)$ such that $p(x)$ is monotonically decreasing on $\left(0, x_{0}\right)$ and monotonically increasing on $\left(x_{0},+\infty\right)$.
Thus, $p(x)_{\min }=p\left(x_{0}\right)=\mathrm{e}^{x_{0}}-4 \ln x_{0}-8+8 \ln 2=\frac{4}{x_{0}}-4\left(\ln 4-x_{0}\right)-8+8 \ln 2=4\left(x_{0}+\frac{1}{x_{0}}\right)-8>0$. In summary, when $x>0$, $f(x)>4 \ln x+8-8 \ln 2$.