Edit 1: I did not know that the question has already been asked. Sorry for inconvenience.
let $x,y \in \mathbb R$ such that $1<x^2-xy+y^2<2$. Prove that $\frac{2}{9}<x^4+y^4<8$.
This question is duplicate of question that was originally asked by Hatim Blilet but the question was put on hold. But I am still interested in question.
$\mathbf{My}$ $\mathbf{approach}$ : I tried proving the inequality by adding the $xy$ term to all sides and squaring all sides of inequality. After that using the maximum value of the quadratic expression obtained on the rightmost part; I got the term 8. But now I have the problem of finding minimum value of leftmost part. The expression I got was $$ -(x^2y^2-2xy-1)\lt x^4+y^4 \lt -(x^2y^2-4xy-4)$$. Also to square both sides we need to prove that $$1\gt xy$$
Second approach : I multiplied all sides by 2 to get $$ 2\lt {(x -y)}^2+x^2+y^2 \lt 4$$.but couldn't proceed further with this method.
Can somebody please tell me how to solve this problem