Continuity Proof: Q about Inequality 0≤(x-y^2)^2

[gajohnson]

Homework Statement

The problem statement and proof can be found here. The proof continues after this, but I only have a question about the beginning of the proof.

Homework Equations

NA

The Attempt at a Solution

My question is simply this:

Every proof I find for this problem starts assuming the inequality $0≤(x-y^2)^2$ and I cannot figure out why. I'm sure this is a simple matter, but any explanation would be appreciated.

Thanks!

 

[jbunniii]

Every proof I find for this problem starts assuming the inequality $0≤(x-y^2)^2$ and I cannot figure out why. I'm sure this is a simple matter, but any explanation would be appreciated.

The square of a real number can't be negative, right?

 

[gajohnson]

The square of a real number can't be negative, right?

Yes, indeed, I see that it works. But was this first step just conjured out of experience and a keen eye, or did something in the problem suggest it specifically?

Thanks!

 

[jbunniii]

Yes, indeed, I see that it works. But was this first step just conjured out of experience and a keen eye, or did something in the problem suggest it specifically?

Probably experience and a keen eye. If you play around with inequalities often enough, you start to recognize things like the fact that the numerator of
$$\frac{xy^2}{x^2 + y^4}$$
can be used to "complete the square" in the denominator, i.e.
$$x^2 + y^4 + 2xy^2 = (x + y^2)^2$$
and
$$x^2 + y^4 - 2xy^2 = (x - y^2)^2$$
If you don't notice this trick, there are usually other ways to find a bound.