<s>[INST] 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.
 [/INST] You want to show that if $x$ and $y$ are nonnegative real numbers with $x+y\leq4$ then
$$f(x,y)=y(x-3)(y-3)-3(4-y),$$
is negative. Taking the derivative with respect to $x$ we get
$$\frac{\partial f}{\partial x}=y(y-3),$$
which is nonzero if $y\neq0$ and $y\neq3$. The derivative is everywhere positive for $y>3$ and so for these values of $y$ the maximum is at $x=4-y$. This shows that for $y>3$ we have
$$f(x,y)\leq f(4-y,y)=y(1-y)(y-3)-3(4-y)=-y^3+4y^2-12.$$
Similarly, if $0<y<3$ then the derivative is negative, and so for these values of $y$ the maximum is at $x=0$. This shows that for $0<y<3$ we have
$$f(x,y)\leq f(0,y)=-3y(y-3)-3(4-y)=-3y^2+6y-12.$$
For $y=0$ and $y=3$ it is clear that $f(x,y)=-3(4-y)$ is negative.
</s>