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?

COMMENT.- In atention to said by @Erik Satie (artistic nickname) we give here another different proof.
We consider $z=a$ fixed. Then the segment with positive $x$ and $y$ of the line $x+y=3-a$ must be “under” the curve in the first quadrant of equation $f(x)=y=\dfrac{2.25-ax}{(1-a)x+a}$ which is equivalent to the proposed inequality for all $x,y$ being $z=a$. Since $f(x)$ is convex in the first quadrant (because $f''(x) \gt0$) and symmetric with respect to the diagonal $y=x$, considering the two fixed points $(x_1,x_1)$ and $(x_2,y_2)$ of the line and the curve respectively, it is enough to verify that $x_1\le x_2$ which reduces the inequality to one of a single variable $a$ easy to verify. You have then to verify the inequality
$$\dfrac{3-a}{2}\le\dfrac{-a+\sqrt{a^2+2.25(1-a)}}{1-a}$$ in which you do have to consider the cases $0\le a\lt1,\space a=1, \space 1\lt a\le3$. Note that for $a=1$ the curve $f(x)$ is a line and the difference $x_1\le x_2$  is evident.
