2. find the highest possible value of the expression

$$
\frac{x y z}{(1+x)(x+y)(y+z)(z+16)}
$$

where $x, y, z$ are positive real numbers.

## 1st solution
Let $A$ be the denominator of the fraction. According to AM-GM

$$
\begin{aligned}
A & =\left(1+\frac{x}{2}+\frac{x}{2}\right)\left(x+\frac{y}{2}+\frac{y}{2}\right)\left(y+\frac{z}{2}+\frac{z}{2}\right)(z+8+8) \\
& \geq 81 \sqrt[3]{x^{2} / 4} \cdot \sqrt[3]{x y^{2} / 4} \cdot \sqrt[3]{y z^{2} / 4} \cdot \sqrt[3]{64 z} \\
& =81 x y z
\end{aligned}
$$

The expression is therefore at most equal to 1/81, equality only applies to $(x, y, z)=(2,4,8)$.

## 2nd solution

Hölder's inequality results in

$$
\begin{aligned}
(1+x)(x+y)(y+z)(z+16) & \geq(\sqrt[4]{1 \cdot x \cdot y \cdot z}+\sqrt[4]{x \cdot y \cdot z \cdot 16})^{4} \\
& =(3 \sqrt[4]{x y z})^{4}=81 x y z
\end{aligned}
$$

The expression is therefore at most equal to 1/81, equality applies to $(x, y, z)=(2,4,8)$.

Alternatively, you can also use CS repeatedly:

$$
\begin{aligned}
(1+x)(y+z)(x+y)(z+16) & \geq(\sqrt{1 \cdot y}+\sqrt{x \cdot z})^{2}(\sqrt{x \cdot z}+\sqrt{y \cdot 16})^{2} \\
& \geq(\sqrt[4]{x y z}+2 \sqrt[4]{x y z})^{4}=81 x y z
\end{aligned}
$$

## 3rd solution

We denote the given expression by $f(x, y, z)$. We now repeatedly use the following

Lemma 1: For $\alpha, \beta>0$ the function

$$
h(t)=\frac{t}{(\alpha+t)(t+\beta)}
$$

reaches its maximum at $t=\sqrt{\alpha \beta}$.

Proof. According to AM-GM

$$
\frac{1}{h(t)}=t+(\alpha+\beta)+\frac{\alpha \beta}{t} \geq 2 \sqrt{t \cdot \frac{\alpha \beta}{t}}+(\alpha+\beta)=(\sqrt{\alpha}+\sqrt{\beta})
$$

with equality if and only if $t=\sqrt{\alpha \beta}$.

If we first fix $y$, $z$, then $x, y$, the following follows from the lemma

$$
f(x, y, z) \leq f(\sqrt{y}, y, z) \leq f(\sqrt{y}, y, 4 \sqrt{y})
$$

with equality if and only if $x=\sqrt{y}$ and $z=4 \sqrt{y}$. A further application of the lemma results in

$$
f(\sqrt{y}, y, 4 \sqrt{y})=\left(\frac{\sqrt{y}}{(1+\sqrt{y})(\sqrt{y}+4)}\right)^{2} \leq\left(\frac{1}{9}\right)^{2}=\frac{1}{81}
$$

with equality exactly when $\sqrt{y}=2$. Overall, the expression is therefore at most equal to $1 / 81$ with equality exactly when $(x, y, z)=(2,4,8)$.