19. Let $x, y, z \geq 0$, and satisfy $x+y+z=1$. Find the maximum value of the following expression $\frac{x-y}{\sqrt{x+y}}+\frac{y-z}{\sqrt{y+z}}+\frac{z-x}{\sqrt{z+x}}$ (Pham Kim Hung)
First, we consider the case where $\min (x, y, z)=0$. Without loss of generality, assume $z=0$, then $x+y=1$, thus
$$\sum_{c y c} \frac{x-y}{\sqrt{x+y}}=\frac{x-y}{\sqrt{x+y}}+\sqrt{y}-\sqrt{x}=x-y+\sqrt{y}-\sqrt{x}=u(v-1)$$

where $u=\sqrt{x}-\sqrt{y}, v=\sqrt{x}+\sqrt{y}, \quad u^{2}+v^{2}=2$
Let $u^{2}(v-1)^{2}=\left(2-v^{2}\right)(v-1)^{2}=f(v)$, we have $f^{\prime}(v)=2(v-1)\left(2+v-2 v^{2}\right)$
and it is easy to see that $\max _{1 \leq v \leq \sqrt{2}} f(v)=f\left(\frac{1+\sqrt{17}}{4}\right)=\frac{71-17 \sqrt{17}}{32}$
Therefore, when $\min (x, y, z)=0$, the maximum value of the expression is $k=\sqrt{\frac{71-17 \sqrt{17}}{32}}$
This result also indicates that if $\min (x, y, z)=0$, then
$$\frac{x-y}{\sqrt{x+y}}+\frac{y-z}{\sqrt{y+z}}+\frac{z-x}{\sqrt{z+x}} \leq k \sqrt{x+y+z}$$

Now we will prove (*) for all non-negative real numbers $x, y, z$ (we do not need the condition $x+y+z=1$, as the inequality is homogeneous). Let $c=\sqrt{x+y}, b=\sqrt{x+z}, a=\sqrt{y+z}$, then the inequality becomes the following form
$$\begin{array}{l}
\frac{b^{2}-a^{2}}{c}+\frac{c^{2}-b^{2}}{a}+\frac{a^{2}-c^{2}}{b} \leq \frac{k}{\sqrt{2}} \cdot \sqrt{a^{2}+b^{2}+c^{2}} \\
\Leftrightarrow(a-b)(b-c)(c-a)\left(\frac{1}{a b}+\frac{1}{b c}+\frac{1}{c z}\right) \leq \frac{k}{\sqrt{2}} \cdot \sqrt{a^{2}+b^{2}+c^{2}}
\end{array}$$