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.
 A: 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 $x=0$
A: If $x>3$ so
$$y\leq4-x<4-3=1,$$ which says the inequality is true.
Thus, we can assume that $x\leq3$.
If $y=0$ or $y\geq3$ we obtain again that the inequality is obvious.
Thus, it's enough to prove our inequality for $0\leq x\leq3$ and $0<y<3$.
Id est, we need to prove that
$$y(3-y)(3-x)\leq3(4-y)$$ or
$$3-x\leq\frac{3(4-y)}{y(3-y)}$$ or
$$x\geq3-\frac{3(4-y)}{y(3-y)}$$ or
$$x\geq\frac{-3(y-2)^2}{y(3-y)},$$ which is obvious again.
A: 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.
