Three, (20 points) For any real numbers $x, y$, we have
$$
|x-2|+|x-4| \geqslant m\left(-y^{2}+2 y\right)
$$

Determine the maximum value of the real number $m$.
Three, by the geometric meaning of absolute value, $|x-2|+|x-4|$ has a minimum value of 2 when $x \in [2,4]$.

And $-y^{2}+2y=-(y-1)^{2}+1$ has a maximum value of 1 when $y=1$.
From the condition, $2 \geqslant m \times 1$, then $m \leqslant 2$.
Therefore, the maximum value of $m$ is 2.