19 Let $x, y, z$ be positive real numbers, such that for some real number $k \geqslant 2$, we have
$$
x y z \leqslant 2,1 \frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}<k .
$$

Find all real numbers $k$, such that $x, y, z$ can be the side lengths of some triangle.
We find that when $x=y=1, z=2$, $x y z \leqslant 2$, and $\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{2}=\frac{9}{4}$. However, 1.1 .2 cannot be the side lengths of a triangle, so $k \leqslant \frac{9}{4}$. We will prove that $\forall k \in\left[\frac{3}{\sqrt[3]{4}}, \frac{9}{4}\right]$ are all solutions that satisfy the problem.
By the problem's condition, $k \geqslant 2$, so the range of $k$ is $\left[2, \frac{9}{4}\right]$.
Assume $x \leqslant y \leqslant z$, and they satisfy $x y z \leqslant 2, \frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}} \geq 0$. Therefore, $p^{2}+p \leqslant 21$. Since $s p \leqslant 2$, then $s^{2} \geqslant 4 p > 4$, which is a contradiction. This indicates that the assumption is not valid. Therefore, $x, y, z$ are the side lengths of a triangle.
When $x=y=z=\sqrt[3]{2}$, $\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}=\frac{3}{\sqrt[3]{4}}<2$, which shows that all real numbers $k$ in the interval $\left[2, \frac{9}{4}\right]$ satisfy the problem.