Let $\displaystyle x,y$ be real. Show that if $\displaystyle x^2 + y^2 > 2$, then $\displaystyle x^4 + y^4 > x^2 + y^2$
If $\displaystyle x = r \cos \theta $ and $\displaystyle y = r \sin \theta$, then you would like to establish that
$\displaystyle r^2 \left( \cos^4 \theta + \sin^4 \theta\right) > 1 \; \text{if}\; r^2 > 2$ or
$\displaystyle
\cos^4 \theta + \sin^4 \theta \ge \frac{1}{2}.
$
Since $\displaystyle \cos^2\theta = 1 - \sin^2 \theta $ then we wish to prove that
$\displaystyle
\sin^4 \theta +(1 - \sin^2 \theta)^2 \ge \frac{1}{2},
$
or
$\displaystyle
2 \sin^4 \theta - 2\sin^2 \theta + 1 \ge \frac{1}{2}
$
Completing the square gives
$\displaystyle
\left( \sin^2 \theta - \frac{1}{2}\right)^2 + \frac{1}{4} \ge \frac{1}{4}
$
which obviously is true.
Assume, without loss of generality, that $\displaystyle x^2\ge 1 \ge y^2$.
$\displaystyle x^2+y^2 > 2 \Leftrightarrow x^2-1 > 1-y^2$,
where $\displaystyle x^2-1 \ge 0$ and $\displaystyle 1-y^2 \le 0$,
so $\displaystyle x^2(x^2-1)>y^2(1-y^2)$, and thus $\displaystyle x^4+y^4 > x^2+y^2$.