Inequality $xy+yz+zx-xyz \leq \frac{9}{4}.$ Currently I try to tackle some olympiad questions:

Let $x, y, z \geq 0$ with $x+y+z=3$. Show that
$$
x y+y z+z x-x y z \leq \frac{9}{4}.
$$
and also find out when the equality holds.


I started by plugging in $z=3-x-y$ on the LHS and got
$$
3y-y^2+3x-x^2-4xy+x^2y+xy^2 = 3y-(y^2+x^2)+3x-4xy+x^2y+xy^2\leq 3y-((y+x)^2)+3x-4xy+x^2y+xy^2
$$
But this got me nowhere.
Then I started again with the left hand side
$$
x y+y z+z x-x y z \Leftrightarrow yz(1-x)+xy+zx
$$
and $x+y+z=3 \Leftrightarrow y+z-2=1-x$ so
$$
yz(y+z-2)+x(y+z)
$$
But this also leaves no idea. Do I have to use a known inequality?

Partial answer with hint :
Many answer are given so it's hard to find a new one so accept my apologize for this partial achivement .
Taking account of the comment due to Thomas Andrew we need to show for $a+c+b=3$,$a,b,c>0$ and $a,b,c\in[0,2]$,$a,b,c\neq 1$:
$$\left|\left(1-a\right)\right|\left|\left(1-b\right)\right|\left|\left(1-c\right)\right|-\frac{1}{4}< 0\tag{I}$$
Using the theorem :
Let $a_1,a_2,a_3,b_1,b_2,b_3>0$ such that :
$$a_1+a_2+a_3\geq b_1+b_2+b_3$$
And $$|a_i-a_j|\leq |b_i-b_j|,1\leq i\leq 3,1\leq j\leq 3,i\neq j$$
Then :
$$a_1a_2a_3\geq b_1b_2b_3$$
Here $a_1=a_2=a_3=\frac{1}{4^{\frac{1}{3}}}$ and $b_1=|1-a|,b_2=|1-c|,b_3=|1-b|$
Supposing we have $(I)$ the inequality is thus proved .
We can sligthly change  $a_1,a_2,a_3$ to get the equality case ($a_1=1,a_2=1/2=a_3$).
