Problem 19. Let \(x, y, z\) be non-negative real numbers such that \(x+y+z=1\). Find the maximum of
\[
\frac{x-y}{\sqrt{x+y}}+\frac{y-z}{\sqrt{y+z}}+\frac{z-x}{\sqrt{z+x}}
\]
Solution. First we consider the problem in the case $\min (x, y, z)=0$. WLOG, suppose that $z=0$, then $x+y=1$ and therefore
$$\sum_{c y c} \frac{x-y}{\sqrt{x+y}}=\frac{x-y}{\sqrt{x+y}}+\sqrt{y}-\sqrt{x}=x-y+\sqrt{y}-\sqrt{x}=u(v-1),$$
where $u=\sqrt{x}-\sqrt{y}, v=\sqrt{x}+\sqrt{y}$ and $u^{2}+v^{2}=2$.
Denote $u^{2}(v-1)^{2}=\left(2-v^{2}\right)(v-1)^{2}=f(v)$ We have $f^{\prime}(v)=2(v-1)\left(2+v-2 v^{2}\right)$ and it's easy to infer that
$$\max _{1 \leq v \leq \sqrt{2}}=f\left(\frac{1+\sqrt{17}}{4}\right)=\frac{71-17 \sqrt{17}}{32}$$

So if $\min (x, y, z)=0$ then the maximum we are looking for is
$$k=\sqrt{\frac{71-17 \sqrt{17}}{32}}$$

This result also shows that if $\min (x, y, z)=0$ then
$$\frac{x-y}{\sqrt{x+y}}+\frac{y-z}{\sqrt{y+z}}+\frac{z-x}{\sqrt{z+x}} \leq k \sqrt{x+y+z}(\star)$$

Now we will prove $(\star)$ for all non-negative real numbers $x, y, z$ (we dismiss the condition $x+y+z=1$ because the inequality is homogeneous). Denote $c=\sqrt{x+y}, b=$ $\sqrt{x+z}, a=\sqrt{y+z}$. The problem can be rewritten in the form
$$\begin{array}{c}
\frac{b^{2}-a^{2}}{c}+\frac{c^{2}-b^{2}}{a}+\frac{a^{2}-c^{2}}{b} \leq \frac{k}{\sqrt{2}} \cdot \sqrt{a^{2}+b^{2}+c^{2}} \Leftrightarrow \\
\Leftrightarrow(a-b)(b-c)(c-a)\left(\frac{1}{a b}+\frac{1}{b c}+\frac{1}{c a}\right) \leq \frac{k}{\sqrt{2}} \cdot \sqrt{a^{2}+b^{2}+c^{2}}(\star \star)
\end{array}$$