11. Solve the inequality $\sqrt{4 x-2}+2 \sqrt{3-x}>\frac{2000}{1999}$.
11. Let $y=\sqrt{4 x-2}+2 \sqrt{3-x}$,

From $\left\{\begin{array}{l}4 x-2 \geqslant 0 \\ 3-x \geqslant 0\end{array}\right.$, we get the domain as $\left[\frac{1}{2}, 3\right]$.
Since $y^{2}=4 x-2+4(3-x)+4 \sqrt{(4 x-2)(3-x)}=10+4 \sqrt{-4 x^{2}+14 x-6} \geqslant 10$,
thus $y \geqslant \sqrt{10}>\frac{2000}{1999}$, meaning the original inequality holds for all values in the domain,
hence the solution set is $\left[\frac{1}{2}, 3\right]$.