For non-negative reals $x,y$ such that $x+y\le 4$ prove that $y(x-3)(y-3) \le 3(4-y)$

Question

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.

Answer 1

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 $x=0$

Answer 2

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.

Answer 3

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.