Prove $x^4+y^4+z^4\leq 1$ is a bounded set and $x^3+y^3+z^3\leq 1$ is an unbounded set

Question

Let $X_1= \left \{ (x,y,z) \in \mathbb{R}^3 \mid x^4+y^4+z^4 \leq 1 \right \}$ and $X_2= \left \{ (x,y,z) \in \mathbb{R}^3 \mid x^3+y^3+z^3\leq 1 \right \}$. Prove $X_1$ is a bounded set and $X_2$ is an unbounded set.

Taking the definiton of bounded set: We say that $X\subset \mathbb{R}^m$ is bounded when it exists some $c<0$ such that $\|x\|\leq c$ for every $x \in X$. I think I might take $c=1$ but I'm not sure what else to do.

Answer 1

The first set is bounded as a consequence of the Cauchy-Schwarz inequality:

$$ x^2+y^2+z^2 \leq \sqrt{3\cdot(x^4+y^4+z^4)} \leq \sqrt{3}.$$ The second set is unbounded since the point $t\cdot(-1,-1,\sqrt[3]{2})$ belongs to it for any $t\in\mathbb{R}$.

Answer 2

In the first case you can take $c=1$ and you have it, because if $\|x\|>1$ than $x\notin X_1$ for Cauchy-Schwartz; in the second case consider that $[(-\infty, 1) \times\{0\} \times \{0\}]\in X_2$ so it can't be bounded.

Answer 3

If $(x,y,z) = (x,-x,0)$, then $x^3+y^3+z^3=0\le 1$ no matter how big $x$ gets, so that's unbounded.

But $x^4+y^4+z^4\le 1$ clearly implies $-1\le x\le 1$ and similarly for $y$ and $z$, so that's bounded.