How big does $r$ need to be, to ensure that $x^4+y^4 >r^2$ for all $x^2+y^2 =r^2$?

Question

Consider the circle $x^2+y^2 = r^2$ for some fixed $r>0$. How big does $r$ need to be, to ensure that $$x^4+y^4 >r^4$$ for all $x^2+y^2 =r^2$? I know that $x^2<x^4$ whenever $|x| >1$, but I'm not sure how to use that here.

Answer 1

By C-S $$(1^2+1^2)(x^4+y^4)\geq(x^2+y^2)^2.$$ Thus, $$x^4+y^4\geq\frac{1}{2}r^4.$$

Answer 2

Hint:Let $x=r\cos{\theta},y=r\sin{\theta}$.

Answer 3

$x^4+y^4=r^4-2r^4\cos^2{\theta}\sin^2{\theta}=r^4(1-\frac{1}{2}\sin^2{2\theta})$

The minimum value of the term in parentheses is $\frac{1}{2}$

So $s^4<\frac{1}{2}r^4$