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?

Assume without loss of generality that $x \le 1$. Then,
$$xy+yz+zx-xyz = x(y+z) + yz(1-x) \le x(y+z) + \left(\frac{y+z}{2}\right)^2(1-x).$$
This is equal to
$$(3-x)\left( x + \frac{(1-x)(3-x)}{2} \right). $$
It is not too difficult to check that this is maximal when $x = 0$ (and $y=z=3/2)$, where it attains a value of $9/4$.
