$x$ and $y$ are two real numbers such that: $1 \leq x^2 + y^2 - xy \leq 2$ Show that: $\frac{2}{9} \leq x^4 + y^4 \leq 8$

I know that the question was posted here somewhere but the technics used to solve it were obscure to me as I didn't learn them yet. And I was given this question in middle school so I think it should be possible using basic math, but I still don't know how. The question is in the title but I'll rewrite it: $$x$$ and $$y$$ are two real numbers such that: $$1 \leq x^2 + y^2 - xy \leq 2$$ Show that: $$\frac{2}{9} \leq x^4 + y^4 \leq 8$$ Any help is really appreciated. Btw the problem is posted here as well as I said How can I prove that $2/9<x^4+y^4<8$?

• If you have seen the identical question on this site before then please add a link to that older question. Commented Oct 30, 2023 at 11:48
• What happens if you write this equation in polar coordinates? Commented Oct 30, 2023 at 11:48
• Oh sorry for that, here's the link: math.stackexchange.com/questions/1145506/… Commented Oct 30, 2023 at 11:56
• And for the polar coordinates, I don't know what these are as I'm in middle school Commented Oct 30, 2023 at 12:01
• Here is another iteration of this question: math.stackexchange.com/questions/2543012/… Commented Oct 30, 2023 at 12:01

Let $$x=u+v$$ and $$y=u-v$$, then the condition gives $$1\leq u^2+3v^2\leq 2$$ and the inequality to be proved is equivalent to $$\frac{1}{9}\leq u^4+v^4+6u^2v^2\leq 4$$ which can be easily proved by $$u^4+v^4+6u^2v^2\geq \frac{(u^2+3v^2)^2}{9}$$ and $$u^4+v^4+6u^2v^2\leq (u^2+3v^2)^2$$