For non-negative reals $x,y$ such that $x+y\le 4$ prove that $y(x-3)(y-3) \le 3(4-y)$ For non-negative reals $x,y$ such that $x+y\le 4$ prove that $y(x-3)(y-3) \le 3(4-y)$
ATTEMPT
We transform the equation into $xy^2+12y-3xy-3y^2\le 12$
I noticed that for $x=0, y=2$ equality is achieved, but I am lost here.

I am interested in elementary solution
$$
\begin{aligned}
(y-2)^{2}&\geq 0\\
y^{2}-4y+4&\geq 0\\
4-y&\geq y(3-y)\\
\\
x&\geq 0\\
3&\geq 3-x
\end{aligned}
$$
Only one of $y(3-y)$ and $3-x$ can be negative, thus $3(4-y)\geq y(3-x)(3-y)$. from the inequalities, equality is when $y=2$ and $x=0$
