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

Question

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 $2/9<x^4+y^4<8$?

Answer 1

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 $u^4+v^4+6u^2v^2\leq (u^2+3v^2)^2$