# Prove that a set in $\mathbb{R}^3$ is bounded

I want to prove that the set $S = \{(x,y,z) \in \mathbb{R}^3: x +yz+z^3-2 \leq 0, z \geq 0, y^2 \leq 2xz \}$

is bounded, but I'm having troubles. I asked myself, how big can $x$ be? And looked at $x + yz + z^3 - 2 \leq 0$. Well by setting $z = 0$ we can see that $x \leq 2$. Then I ask myself, how small can $x$ be? Well $x \geq \frac{y^2}{2z}$. But that's all I've got.

EDIT

As pointed out, the set isnt bounded. The problem comes from:

Find the minimum of $y^2 - 2xz$ subject to $x+yz+x^3 \leq 2, \ z \geq 0$. And to prove existence the prof. says "Take the point $(0,0,0)$, add the constraint $y^2 \leq 2xz$ and show that the new set is bounded (a bit tricky)." But I'm clearly missing something?

• Isn't $\{x\le 2, y=0,z=0\}$ contained in $S?$ – zhw. Apr 10 '18 at 18:59
• Yes it is. @zhw. – Olba12 Apr 10 '18 at 19:08
• So $x$ is bounded between $0\leq x \leq 2$. – Olba12 Apr 10 '18 at 19:08
• Where did your last comment come from? – zhw. Apr 10 '18 at 19:18
• I was thinking that $x$ needed to be positive from $y^2 \leq 2xz$ I forgot about the the possibility of choosing $z=0$ @zhw. – Olba12 Apr 10 '18 at 19:19

As zhw's comment shows, the set is not bounded. Points with $y=0,z=0,$ and $x$ large and negative are in the set.