Three. (20 points) Let real numbers $a, m$ satisfy
$$
a \leqslant 1, 0 < m \leqslant 2 \sqrt{3} \text{, }
$$

The function $f(x)=\frac{a m x - m x^{2}}{a + a(1-a)^{2} m^{2}}(x \in (0, a))$. If there exist $a, m, x$ such that $f(x) \geqslant \frac{\sqrt{3}}{2}$, find all real values of $x$.
$$
\begin{array}{l}
\text { Three, when } x \in(0, a) \text {, } \\
a m x - m x^{2} \\
= -m\left(x - \frac{a}{2}\right)^{2} + \frac{m a^{2}}{4} \leqslant \frac{m a^{2}}{4} .
\end{array}
$$

The equality holds if and only if $x = \frac{a}{2}$.
$$
\begin{array}{l}
\text { Therefore, } \frac{\sqrt{3}}{2} \leqslant \frac{a m x - m x^{2}}{a + a(1-a)^{2} m^{2}} \\
\leqslant \frac{\frac{a^{2}}{4} m}{a + a(1-a)^{2} m^{2}} \\
= \frac{a m}{4\left[1 + (1-a)^{2} m^{2}\right]} \\
\leqslant \frac{a m}{4} \leqslant \frac{m}{4} \leqslant \frac{\sqrt{3}}{2} .
\end{array}
$$

The equality holds if and only if $x = \frac{a}{2}$ and $a = 1, m = 2 \sqrt{3}$. Therefore, $x = \frac{1}{2}$.