Minimizing cubic over elliptic disk without using derivatives Given $x, y$ satisfy $(x-4)^2 + (y-4)^2 + 2xy \leq 32$. Find the minimum value of:
$$P = x^3 + y^3 +3(xy-1)(x+y-2)$$
My attempt:
From $(x-4)^2 + (y-4)^2 + 2xy \leq 32$, I get:
$$(x+y)^2 - 8(x+y) \leq 0$$
$$\implies 0 \leq x+y \leq 8$$
From $P = x^3 + y^3 +3(xy-1)(x+y-2)$, I get:
$$P = (x+y)^3 - 6xy - 3(x+y) + 6 \geq (x+y)^3 - 3(x+y)^2 - 3(x+y) + 6$$
I can use derivatives to find the minimum value but my teacher told me to try to solve this without derivatives. Is there a way to do that?
Edit: There was a mistake in my solving. I was using AM-GM inequality to derive this:
$$- 6xy \geq - \frac{3(x+y)^2}{2}$$
But I forgot that $x$ and $y$ are both real and can be negative.
So now I fixed it. Instead of $\frac{3(x+y)^2}{2}$, it should be $3(x+y)^2$ since $2xy \leq (x + y)^2$
 A: Yes, we can.
Let $x+y=2u$ and $xy=v^2$, where $v^2$ can be negative.
Thus, the condition gives $$(x+y)^2-8(x+y)\leq0$$ or
$$0\leq u\leq4.$$
Now, since $u^2\geq v^2$ it's $(x-y)^2\geq0$, we obtain: $$P=8u^3-6uv^2+3(v^2-1)(2u-2)=8u^3-6v^2-6u+6\geq$$
$$\geq8u^3-6u^2-6u+6=\left(2u-\frac{1}{2}\right)^3-\frac{3}{2}u+\frac{1}{8}-6u+6=$$
$$=\left(2u-\frac{1}{2}\right)^3-\frac{15}{4}\left(2u-\frac{1}{2}\right)+\frac{17}{4}=$$
$$=\left(2u-\frac{1}{2}\right)^3+2\cdot\frac{\sqrt{125}}{8}-\frac{15}{4}\left(2u-\frac{1}{2}\right)+\frac{17}{4}-\frac{5\sqrt5}{4}\geq$$
$$\geq 3\sqrt[3]{\left(2u-\frac{1}{2}\right)^3\cdot\left(\frac{\sqrt{125}}{8}\right)^2}-\frac{15}{4}\left(2u-\frac{1}{2}\right)+\frac{17}{4}-\frac{5\sqrt5}{4}=\frac{17-5\sqrt5}{4}.$$
We used the following inequality.
$$\left(2u-\frac{1}{2}\right)^3+\frac{\sqrt{125}}{4}\geq\frac{15}{4}\left(2u-\frac{1}{2}\right),$$ which is
$$\left(u-\frac{1+\sqrt5}{4}\right)^2\left(u+\frac{2\sqrt5-1}{4}\right)\geq0,$$ which is true for $u\geq0.$
The equality occurs for $x=y=\frac{1+\sqrt5}{4},$ which says that we got a mninimal value.
