For $x=2, y=3$ and $z=6$ the equality holds.

After the substitutions $x=2-u, y=3-v$ with $u \in[0,2), v \in[0,3)$, we obtain that $z=6+u+v$ and the required inequality becomes

$$(2-u)(3-v)(6+u+v) \leqslant 36$$

We shall need the following lemma.

Lemma. If real numbers $a$ and $b$ satisfy the relations $0 < a \leq b$ and $y \geq 0$.

The equality in (2) holds if $y=0$. The lemma is proved.

By using the lemma we can write the following inequalities:

$$\begin{gathered} \frac{6}{6+u} \geqslant \frac{2-u}{2} \\ \frac{6}{6+v} \geqslant \frac{3-v}{3} \\ \frac{6+u}{6+u+v} \geqslant \frac{6}{6+v} \end{gathered}$$

By multiplying the inequalities (3), (4) and (5) we obtain:

$$\begin{gathered} \frac{6 \cdot 6 \cdot(6+u)}{(6+u)(6+v)(6+u+v)} \geqslant \frac{6(2-u)(3-v)}{2 \cdot 3(6+v)} \Leftrightarrow \\ (2-u)(3-v)(6+u+v) \leqslant 2 \cdot 3 \cdot 6=36 \Leftrightarrow \quad(1) \end{gathered}$$

By virtue of lemma, the equality holds if and only if $u=v=0$.

Alternative solution. With the same substitutions write the inequality as

$$(6-u-v)(6+u+v)+(u v-2 u-v)(6+u+v) \leq 36$$

As the first product on the lefthand side is $36-(u+v)^{2} \leq 36$, it is enough to prove that the second product is nonpositive. This comes easily from $|u-1| \leq 1$, $|v-2| \leq 2$ and $u v-2 u-v=(u-1)(v-2)-2$, which implies $u v-v-2 u \leq 0$.

## Geometry