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?

For $x=y=z=1$ we obtain on the L.H.S. the value two. Which is far away from the R.H.S. $9/4$. What about showing a stronger inequality under the given constraints?
(Of course, we are slightly changing the "weakest" term...)
$$
\boxed{\qquad
xy + yz + zx - \color{blue}{\frac 34} xyz \ \le\ \frac 94
\qquad}
$$
It is a good idea to switch to the homogenized version, we will ultimately find an inequality involving an algebraic expression of degree three for three variables, there are not too many of this kind...
$$
\tag{$1$}
\frac 13(x+y+z)(xy+xz+zx) - \frac 34xyz\le\frac 94\cdot\frac 1{27}(x+y+z)^3\ .
$$
After moving everything on the R.H.S., expanding, simplifying and getting rid of the denominator $12$, we obtain:
$$
\tag{$2$}
0\le x^3+y^3+z^3 - x^2y-xy^2-y^2 z-yz^2-z^2x-zx^2+ 3xyz\ ,
$$
and for sure, this is Schur.
$\square$

Note that expanding
$(1)$ is not a complicated dirty step, that would eliminate this solution from an olympiad priced solution, this step is easily done. We move all terms on the R.H.S. of $(1)$, multiply by $12$ to get
$$
\begin{aligned}
&(x+y+z)^3 -4(x+y+z)(xy+yz+zx) +9xyz
\\[3mm]
&\qquad =
\sum\left(\binom 3{3 \,0\,0}x^3 + \binom 3{2 \,1\,0}\color{brown}{x^2y} + \binom 3{1 \,1\,1}xyz \right)
- 4\left(\sum_{\text{cyclic}}x(\color{brown}{xy}+yz+\color{brown}{zx})\right)
+ 9xyz
\\[2mm]
&\qquad=
\sum_{\text{cyclic}} x^3 
+ \sum_{\text{cyclic}}(3-4)\color{brown}{(x^2y + x^2z)} 
+ (6 - 12 + 9)xyz
\end{aligned}
$$
which is the expression in the R.H.S. of $(2)$.
