Compare two functions in neighborhood of zero

Question

Basically, I have two functions: $$ f(x, y) = \sqrt{(x-y^2)^2 + x^4} $$ and $$ g(x, y) = | (x-y^2)^3 | $$ I need to compare them in a punctured neighborhood of zero. I am a bit stuck here.

Edit: compare in sense that there exists a punctured neighborhood of zero where for all $x$ and $y$ from that neighborhood one function is greater than the other.

I.e. I want to prove that $$ \exists O(0,0):\quad \forall x, y \in O(0,0) \quad |f(x, y)| > |g(x,y)| $$

Answer 1

Let $ w = x - y^2 $.

If $ w > 0 $ and $w < 1$:

$$ f = \sqrt{w^2 + x^4} \ge \sqrt{w^2} = w $$

$$ g = w^3 < w $$

Therefore:

$$ f \ge w > g $$

If w < 0 and | w | < 1

$$ f = \sqrt{w^2 + x^4} \ge \sqrt{w^2} = -w $$

$$ g = | w^3 | < -w $$

Therefore:

$$ f \ge -w > g $$

If $ w = 0 $ and $ x \ne 0 $

$$ f = \sqrt{x^4} = x^2 > 0 $$

$$ g = w^3 = 0 $$

Therefore:

$$ f > 0 = g $$

$ |w| < 1 $ is a pretty big neighborhood, relatively speaking.

Hope this helps.

Ced

Answer 2

If $|(x-y^2)|\le 1$ then $g(x,y)\le \sqrt{(x-y^2)^2}\le f(x,y)$. Can you find $O(0,0)$?