Four, (20 points) Given $f(x)=2 x-\frac{2}{x^{2}}+\frac{a}{x}$, where the constant $a \in(0,4]$. Find all real numbers $k$, such that for any $x_{1} 、 x_{2} \in \mathbf{R}_{+}$, it always holds that
$$
\left|f\left(x_{1}\right)-f\left(x_{2}\right)\right| \geqslant k\left|x_{1}-x_{2}\right| .
$$
When $x_{1}=x_{2}$, $k$ is arbitrary.
When $x_{1} \neq x_{2}$, the inequality becomes $\left|\frac{f\left(x_{1}\right)-f\left(x_{2}\right)}{x_{1}-x_{2}}\right| \geqslant k$.
$$
\begin{array}{l}
\text { Since } \frac{f\left(x_{1}\right)-f\left(x_{2}\right)}{x_{1}-x_{2}} \\
=\frac{1}{x_{1}-x_{2}}\left[2\left(x_{1}-x_{2}\right)-\left(\frac{2}{x_{1}^{2}}-\frac{2}{x_{2}^{2}}\right)+\left(\frac{a}{x_{1}}-\frac{a}{x_{2}}\right)\right] \\
=2+\frac{2\left(x_{1}+x_{2}\right)}{x_{1}^{2} x_{2}^{2}}-\frac{a}{x_{1} x_{2}} \\
>2+\frac{4 \sqrt{x_{1} x_{2}}}{x_{1}^{2} x_{2}^{2}}-\frac{a}{x_{1} x_{2}} \\
=\left(2-\frac{a^{3}}{108}\right)+\frac{a^{3}}{108}+\frac{2 \sqrt{x_{1} x_{2}}}{x_{1}^{2} x_{2}^{2}}+\frac{2 \sqrt{x_{1} x_{2}}}{x_{1}^{2} x_{2}^{2}}-\frac{a}{x_{1} x_{2}} \\
\geqslant\left(2-\frac{a^{3}}{108}\right)+3 \sqrt[3]{\frac{a}{108}\left(\frac{2 \sqrt{x_{1} x_{2}}}{x_{1}^{2} x_{2}^{2}}\right)^{2}}-\frac{a}{x_{1} x_{2}} \\
=2-\frac{a^{3}}{108}, \\
\text { hence } \frac{f\left(x_{1}\right)-f\left(x_{2}\right)}{x_{1}-x_{2}}>2-\frac{a^{3}}{108} .
\end{array}
$$

When $x_{1} \neq x_{2}$, and both tend to $\frac{6}{a}$, we have
$$
\frac{f\left(x_{1}\right)-f\left(x_{2}\right)}{x_{1}-x_{2}} \rightarrow 2-\frac{a^{3}}{108} \text {. }
$$

Therefore, the set of $k$ is $\left(-\infty, 2-\frac{a^{3}}{108}\right]$.