How can I prove that $2/9

Question

$x$ and $y$ are real numbers.

Given that $1<x^2-xy+y^2<2$, how can I show that $\frac 29<x^4+y^4<8$ ?

Then can I use that to prove that for any natural number $n>3$

$$x^{2^n}+y^{2^n}>\frac 2{3^{2^n}} \text{?}$$

Answer 1

1) to show $x^4+y^4 < 8$, you already have a good solution by squaring $x^2+y^2<2+xy$.

2) By AM-GM and Cauchy Schwarz, we get

$$\left(x^4+y^4+\frac{x^4+y^4}2\right)(3) \ge (x^4+y^4+x^2y^2)(1^2+1^2+(-1)^2) \ge (x^2+y^2-xy)^2 > 1$$ hence $x^4+y^4 > \frac29$

For the more general case, you could use Power Means in addition to the above, i.e. for $n>2$: $$\sqrt[2^n]{\frac{x^{2^n}+y^{2^n}}2} \ge \sqrt[4]{\frac{x^4+y^4}2} > \frac1{\sqrt3} \implies x^{2^n}+y^{2^n} > \frac2{3^{2^{n-1}}}$$ which is a tighter bound than your inequality.

Answer 2

$x^2 - xy + y^2 = 1$ is an ellipse. to find out the minor and major axis, we change variables $$x = \xi \cos t - \eta \sin t, y = \xi \sin t + \eta \cos t$$ where $t$ will be determined later.

the ellipse in $\xi, \eta$ coordinates is $$\xi^2(\cos^2 t -\sin t \cos t+\sin^2 t) + \xi \eta(-2\cos t \sin t+\sin^2 t - \cos^2 t+2\sin t \cos t) + \eta^2(\sin ^2 t +\sin t \cos t +\cos ^2 t) =1 $$

set $t = \frac{\pi}{4},$ you get $$\frac12\xi ^2+\frac32\eta^2= 1. $$

so the major and minor axis of the ellipse $x^2 - xy + y^2 = 1$ are $\sqrt 2$ and $ \sqrt { \frac23}.$

therefore by symmetry the minimum of$x^4 + y^4$ on the hyperbola $x^2 -xy + y^2 = 1$ is achieved on the minor axis $x = -\sqrt\frac13, y = \sqrt \frac 13$ and the minimum value is $\dfrac{2}{9}.$

therefore by symmetry the minimum of$x^4 + y^4$ on the hyperbola $x^2 -xy + y^2 = 2$ is achieved on the major axis $x = \sqrt 2, y = \sqrt 2$ and the maximum value is $8.$

we have established $$ \dfrac{2}{9} \le x^4 + y^4 \le 8 \text { on } 1 \le x^2 - xy + y^2 \le 2.$$

Answer 3

$0\leq x^2+y^2<2+xy$. Squaring it, you get $x^4+y^4+2x^2y^2<4+4xy+x^2y^2\Rightarrow x^4+y^4<4+4xy-x^2y^2=8-(4+x^2y^2-4xy)=8-(2-xy)^2\leq 8$