Since $\angle MAB = \angle CAB - \angle CAM$, according to the sine addition theorem we have
$$\sin(\angle MAB) = \sin(\angle CAB) \cos(\angle CAM) - \cos(\angle CAB) \sin(\angle CAM).$$

Let $|BC| = a$, $|CA| = b$, $|AB| = c$ and $|AM| = t$. From right triangles $ABC$ and $AMC$ we get
$$\sin(\angle MAB) = \frac{a}{c} \cdot \frac{b}{t} - \frac{b}{c} \cdot \frac{a/2}{t} = \frac{ab}{2ct}.$$

Since $c^2 = a^2 + b^2$ and $t^2 = \frac{a^2}{4} + b^2$, that is equivalent to
$$9a^2b^2 \le (a^2 + b^2)(a^2 + 4b^2)$$
i.e. $(a^2 - 2b^2)^2 \ge 0$. Now it is clear that the given inequality holds in every right triangle. The equality is achieved when $a = b\sqrt{2}$.