Find the set of values taken by the function $f:] 0,+\infty\left[{ }^{3} \mapsto\right.$ $\mathbb{R}$ such that $f(x, y, z)=\frac{x \sqrt{y}+y \sqrt{z}+z \sqrt{x}}{\sqrt{(x+y)(y+z)(z+x)}}$.
First, note that $f$ only takes strictly positive values. Moreover, if $\varepsilon \rightarrow 0$, then

$$
f\left(\varepsilon^{-1}, \varepsilon^{2}, 1\right)=\frac{1+\varepsilon^{2}+\varepsilon^{-1 / 2}}{\sqrt{\left(\varepsilon^{-1}+\varepsilon^{2}\right)\left(\varepsilon^{2}+1\right)\left(1+\varepsilon^{-1}\right)}} \sim \frac{\varepsilon^{-1 / 2}}{\varepsilon^{-1}}=\varepsilon^{1 / 2} \rightarrow 0
$$

so that $\inf f(] 0,+\infty\left[{ }^{3}\right)=0$.

On the other hand, we know that $\sqrt{x y} \leq \frac{x+y}{2}$, that $\sqrt{y z} \leq \frac{y+z}{2}$, and that $\sqrt{z x} \leq \frac{z+x}{2}$. This is why

$$
\begin{aligned}
(x \sqrt{y}+y \sqrt{z}+z \sqrt{x})^{2} & =\left(x^{2} y+y^{2} z+z^{2} x\right)+2(x y \sqrt{y z}+y z \sqrt{z x}+z x \sqrt{x y}) \\
& \leq\left(x^{2} y+y^{2} z+z^{2} x\right)+2\left(x y \frac{y+z}{2}+y z \frac{z+x}{2}+z x \frac{x+y}{2}\right) \\
& \leq(x+y)(y+z)(z+x)+x y z \\
& \leq(x+y)(y+z)(z+x)+\frac{x+y}{2} \frac{y+z}{2} \frac{z+x}{2}=\frac{9}{8}(x+y)(y+z)(z+x)
\end{aligned}
$$

so that $f(x, y, z) \leq \frac{3}{2 \sqrt{2}}$, with equality if $x=y=z=1$.

We therefore deduce that $\left.\left.f(] 0,+\infty\left[{ }^{3}\right)=\right] 0, \frac{3}{2 \sqrt{2}}\right]$.