Solution:

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 $0 < b \leq a$, then for every real number $y \in [0, b)$ the inequality
$$\frac{a}{a+y} \geqslant \frac{b-y}{b}$$
holds.

Proof of the lemma. The inequality (2) is equivalent to
$$a b \geq a b - a y + b y - y^{2} \Leftrightarrow y^{2} + (a-b) y \geq 0$$
The last inequality is true, because $a \geq b > 0$ 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 $u = v = 0$.

Solution:

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 left-hand 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 $u v-v-2 u \leq 0$.