What does $x^2 \geq y^2$ imply? What does $x^2 \geq y^2$ imply?
I'm struggling a little on the question I thought of when working another problem.
Do we simply have that if $x^2 \geq y^2$ then $x \geq y$ and nothing else?
Or can we relate $-y$ to $x$ some how. I guess my confusion stems from the fact that $\sqrt {x^2}$ returns only positive $x$.
Thanks.
 A: You could take the square root of both sides, keeping in mind that $\sqrt{a^2} = |a|$, so:
$$x^2 \ge y^2 \Leftrightarrow |x| \ge |y| $$
What does this mean for $x$ and $y$?
Or, rewrite:
$$x^2 \ge y^2 \Leftrightarrow x^2 - y^2 \ge 0 \Leftrightarrow (x-y)(x+y) \ge 0$$
Can you take it from here?
You could go for a graphical solution as well, drawing the lines $y = x$ and $y = -x$.
A: If you apply any monotonically increasing function to both sides of an inequality, then the inequality still holds: $x \geq y$ implies $f(x) \geq f(y)$. (Actually, this is the definition of an increasing function).
Since $f(x)=\sqrt{x}$ is an increasing function, $x^2 \geq y^2$ implies that $\sqrt{x^2} \geq \sqrt{y^2}$. 
You are right to worry about positives and negatives. $\sqrt{x^2}=|x|$ (the absolute value of $x$). Thus you have $|x| \geq |y|$. 
Anytime you have $|A| \leq B$, then $-B \leq A \leq B$ (Why? $A \geq 0$ yields: $A = |A| \leq B$ and $A <0$ yields: $-A = |A| \leq B$ so that $A \geq -B$).
Using this, you get: $-|x| \leq y \leq |x|$.
This is the region above the negative absolute value function and below the absolute value function (looks like a bow tie).
A: The square of a number is greater than that of another if it is bigger when dropping the signs. In orther words,
$x^2\ge y^2 \Leftrightarrow |x|\ge |y|$.
A: If $x\geq 0$, then 
$$x\geq y, x\leq -y\qquad\Rightarrow -x\leq y\leq x$$
If $x\leq 0$ then
$$-x\geq y, -x\leq -y\qquad \Rightarrow x\leq y\leq -x$$

