10. 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 $x_{1}, x_{2} \in \mathbf{R}^{+}$, it always holds that $\left|f\left(x_{1}\right)-f\left(x_{2}\right)\right|>k\left|x_{1}-x_{2}\right|$.
10. Proof: When $x_{1}=x_{2}$, $k$ is arbitrary; when $x_{1} \neq x_{2}$, the inequality becomes $\left|\frac{f\left(x_{1}\right)-f\left(x_{2}\right)}{x_{1}-x_{2}}\right|>k$. Since
$$
\begin{array}{l}
\frac{f\left(x_{1}\right)-f\left(x_{2}\right)}{x_{1}-x_{2}}=\frac{1}{x_{1}-x_{2}}\left[2\left(x_{1}-x_{2}\right)-\left(\frac{2}{x_{1}^{2}}-\frac{2}{x_{2}^{2}}\right)+\left(\frac{a}{x_{1}}-\frac{a}{x_{2}}\right)\right] \\
=2+\frac{2\left(x_{1}+x_{2}\right)}{x_{1}^{2} x_{2}^{2}}-\frac{a}{x_{1} x_{2}}>2+\frac{4 \sqrt{x_{1} x_{2}}}{x_{1}^{2} x_{2}^{2}}-\frac{a}{x_{1} x_{2}}=\left(2-\frac{a^{3}}{108}\right)+\frac{a^{3}}{108}+\frac{2 \sqrt{x_{1} x_{2}}}{x_{1}^{2} x_{2}^{2}}+\frac{2 \sqrt{x_{1} x_{2}}}{x_{1}^{2} x_{2}^{2}}-\frac{a}{x_{1} x_{2}} \\
\geqslant\left(2-\frac{a^{3}}{108}\right)+3 \cdot \sqrt[3]{\frac{a^{3}}{108} \cdot\left(\frac{2 \sqrt{x_{1} x_{2}}}{x_{1}^{2} x_{2}^{2}}\right)^{2}}-\frac{a}{x_{1} x_{2}}=2-\frac{a^{3}}{108},
\end{array}
$$

thus $\frac{f\left(x_{1}\right)-f\left(x_{2}\right)}{x_{1}-x_{2}}>2-\frac{a^{3}}{108}$. When $x_{1} \neq x_{2}$ and both approach $\frac{6}{a}$, we have $\frac{f\left(x_{1}\right)-f\left(x_{2}\right)}{x_{1}-x_{2}} \rightarrow 2-\frac{a^{3}}{108}$. Therefore, the set of $k$ is $\left(-\infty, 2-\frac{a^{3}}{108}\right]$.