Show that Set in $M:=\{x\in \Bbb R^3 : x_1^2\ge2(x_2^3+x_3^3) \}$ is closed I have to show this regarding the Euclidean metric.

I've already shown that it isn't bounded by showing that the $d(x,y)\:\forall x,y \in M$ isn't bounded.

I know that in order to show the closedness i have to show that the complement is open, but how do I show that $$M:=\{x\in \Bbb R^3 : x_1^2\lt 2(x_2^3+x_3^3) \}$$ is open? (I know that a set is open if the $\epsilon$-ball around every element is also in the set but how do I show this in this particular case)

I have to determine whether the set is closed and whether it's bounded for the following set too, but I don't have any ideas how to show either:
$$M:=\{x\in \Bbb R^3 : x_1 \ge -1, \:x_2\lt2, ||x||_2\le\sqrt5\}$$
Thanks in advance
Let $f(x_1,x_2,x_3)= x_1^2-2(x_2^3+x_3^3)$, which is a polynomial, hence continuous. Then the set in question is $f^{-1}([0,\infty))$ which is closed, hence by the definition of continuity, the inverse image of this closed set is closed.
You need to show that for every $x\in M$ you can find an open ball that is still in $M$. The open Ball definition is :
\begin{align}
B_r(x) = \{ y\in \mathbb{R^3} | \sqrt{\sum_{i=1}^3 (y_i-x_i)^2} &lt; r)\}
\end{align}

So you have an inequality for all y in the ball. You have an inequality for all x int the set $M$. Combining the two if you can find a $r$ that satisfy both, you are done !